Algebra

Ikkinchi va yuqori darajali tenglamalar sistemasi

murakkab 180 daqiqa nochiziqli sistemanonlinear systemsubstitutioneliminationordered pairsymmetric systemx+yxysum and productdifference of squarescubic identityzero productreciprocal systemcyclic systemsystem verification

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Nochiziqli tenglamalar sistemalarida bitta tenglamani yechish yetarli emas: yechim ordered pair yoki tuple bo‘lib, sistemadagi har bir tenglamani bir vaqtda qanoatlantirishi kerak. Substitution, elimination, factorization va symmetric invariants yordamida yuqori darajali sistemalar bir o‘zgaruvchili tenglamaga tushiriladi. Eng katta xavf — factorlashdan chiqqan branchlardan birini tashlab yuborish, ordered pair tartibini aralashtirish yoki kandidatni barcha original tenglamalarda tekshirmaslik. Bu mavzu algebra, analitik geometriya, parametrli sistemalar va optimallashtirishga tayanch beradi.

O'quv maqsadlari

Nochiziqli tenglamalar sistemasini ta’riflash
Ordered pair va ordered tuple yechim tushunchasini qo‘llash
Har bir kandidatni sistemaning barcha tenglamalarida tekshirish
Substitution usulida bir o‘zgaruvchini yo‘qotish
Elimination usulida x² yoki y² kabi hadlarni yo‘qotish
Line + quadratic sistemani bir o‘zgaruvchili quadraticga keltirish
x+y=S va xy=P symmetric invariantsdan foydalanish
x²+y²=S²−2P identikligini sistemalarda qo‘llash
x²−y²=(x−y)(x+y) orqali sistemani linear invariantlarga tushirish
x³−y³ va x³+y³ factor identiklaridan foydalanish
Symmetric sistemani x=y va boshqa factor branchlariga ajratish
x²+y=k va y²+x=k tipida barcha branchlarni topish
x(x+y)=A, y(x+y)=B sistemalarini umumiy factor orqali yechish
Reciprocal sistemalarda domain restrictionlarni saqlash
Kasrli sistemalarda denominator nollarini chiqarish
Zero-product sistemalarda branch tree tuzish
A²+B²=0 va yuqori juft darajalar yig‘indisi kriteriyini qo‘llash
Cyclic uch o‘zgaruvchili sistemalarda simmetriya va minimum argumentlarini tanish
Geometrik masalalarni x+y,xy sistemasiga aylantirish
Source-bankdagi missing variable, missing instruction va wrong marked-answer holatlarini independent QA bilan ajratish

Nochiziqli sistema kamida bitta tenglamasi chiziqli bo‘lmagan tenglamalar to‘plamidir. Yechim — bitta son emas, masalan (x,y) ordered pair bo‘lib, u sistemadagi barcha tenglamalarni bir vaqtda rost qilishi kerak. Grafik nuqtayi nazardan yechimlar egri chiziqlarning kesishish nuqtalaridir. Asosiy algebraik usullar — substitution va elimination. Lekin bankning yuqori qatlamida bundan kuchliroq struktura bor: symmetric sistemalarda S=x+y va P=xy; x³−y³=(x−y)(x²+xy+y²); zero-product tenglamalarda har bir factor alohida branch; reciprocal sistemalarda esa denominator domaini. Branchlar soni ko‘paygan sari final verification muhimlashadi. Canonical yechim algoritmi: domainni aniqlash → strukturani tanish → branch yoki bir o‘zgaruvchili tenglama hosil qilish → barcha kandidat ordered pairlarni tiklash → har bir pairni barcha original tenglamalarda tekshirish.

Ta'riflar

Nochiziqli sistema · Nonlinear system · Нелинейная система

Kamida bitta tenglamasi o‘zgaruvchilarga nisbatan chiziqli bo‘lmagan tenglamalar sistemasi.

Kvadrat, ko‘paytma, kasr yoki yuqori daraja qatnashgan sistema.

Misol: {x+y=5, x²+y²=13}.

Bu emas: {x+y=5, 2x−y=1} to‘liq chiziqli sistema.

💡 Yechimlar grafiklarda kesishish nuqtalariga mos keladi.

Ordered pair · Ordered pair · Упорядоченная пара

(x,y) ko‘rinishidagi tartibli juftlik.

x va y rollari almashtirilmaydi.

Misol: (2,3) da x=2,y=3.

Bu emas: (2,3) va (3,2) odatda bir xil yechim emas.

💡 Symmetric sistemada ikkalasi ham yechim bo‘lishi mumkin.

Sistema yechimi · Solution of a system · Решение системы

Sistemadagi har bir tenglamani bir vaqtda qanoatlantiradigan ordered tuple.

Bir tenglamani rost qilish yetarli emas.

Misol: (2,1) ni ikki tenglamaning ikkalasiga ham qo‘yib tekshirish.

Bu emas: Faqat birinchi tenglamani qanoatlantirgan pair final solution emas.

💡 Final QAning asosiy kriteriyi.

Substitution usuli · Substitution method · Метод подстановки

Bir tenglamadan variable’ni ifodalab boshqa tenglamaga qo‘yish usuli.

Ikki variable sistemani bir variable equationga tushiradi.

Misol: y=x+1 ni x²+y²=25 ga qo‘yish.

Bu emas: Variable’ni ifodalamay turib tasodifiy qiymat qo‘yish substitution emas.

💡 Line + quadratic sistemalarda juda samarali.

Elimination usuli · Elimination method · Метод исключения

Tenglamalarni algebraik qo‘shish/ayirish orqali variable yoki nonlinear hadni yo‘qotish.

x² hadlari bir xil bo‘lsa tenglamalarni ayirib y ni topish mumkin.

Misol: {x²+y²=4, x²−y=4}.

Bu emas: Tenglamalarni shartsiz bo‘lib yuborish elimination emas.

💡 Nonlinear sistemalarda ham linear sistemadagi prinsip ishlaydi.

Symmetric sistema · Symmetric system · Симметрическая система

x va y almashtirilganda sistema o‘zgarmaydigan sistema.

Yechim (a,b) bo‘lsa ko‘pincha (b,a) ham yechim.

Misol: {x+y=6,xy=8}.

Bu emas: {x−y=3,xy=10} symmetric emas.

💡 S=x+y va P=xy tabiiy koordinatalar.

Symmetric invariants · Symmetric invariants · Симметрические инварианты

x,y almashganda o‘zgarmaydigan S=x+y, P=xy kabi kattaliklar.

Individual x,y o‘rniga sum/product bilan ishlash.

Misol: x²+y²=S²−2P.

Bu emas: x−y invariant emas.

💡 Viyet bilan bevosita bog‘liq.

Branch · Solution branch · Ветвь решения

Factorization yoki square-root/parity qadamidan kelib chiqadigan alohida algebraik holat.

Har branch mustaqil yechilib tekshiriladi.

Misol: (x−y)(x+y−1)=0 → x=y yoki x+y=1.

Bu emas: Faqat bir factorni nolga tenglash branchlarni to‘liq bermaydi.

💡 Bank 6954 branchni tashlab yuborish xatosini ko‘rsatadi.

Zero-product sistema · Zero-product system · Система с нулевым произведением

Kamida bir tenglamasi factorlar ko‘paytmasi nol ko‘rinishida bo‘lgan sistema.

Har product equation bir nechta branchga ajraladi.

Misol: (x−5)(y−3)=0.

Bu emas: x−5+y−3=0 zero-product qoidasi emas.

💡 Branch tree bilan yechish qulay.

Domain restriction · Domain restriction · Ограничение области определения

Kasr, ildiz yoki boshqa aniqlanish shartlaridan keladigan taqiqlar.

Kandidat pair denominatorni nol qilsa solution emas.

Misol: 1/x qatnashsa x≠0.

Bu emas: Polynomial sistemada avtomatik x≠0 deb olish noto‘g‘ri.

💡 Ratsional sistemalarda oldindan yoziladi.

Reciprocal system · Reciprocal system · Система с обратными величинами

1/x,1/y yoki x/y kabi reciprocal ratios qatnashgan sistema.

Domain + sum/product invariantlari ko‘p ishlatiladi.

Misol: x+y=7, 1/x+1/y=7/12.

Bu emas: x+y=7,xy=12 reciprocal sistema emas.

💡 1/x+1/y=(x+y)/(xy).

Difference-of-squares reduction · Difference-of-squares reduction · Разность квадратов

x²−y² ni (x−y)(x+y) ga factorlab sistema invariantlarini olish usuli.

Ayirma va yig‘indi bir-biriga bog‘lanadi.

Misol: x−y=3, x²−y²=21 → x+y=7.

Bu emas: x²+y² factorlanmaydi.

💡 Ko‘plab bank savollarida bir qadamlik reduction.

Cubic factor identity · Cubic factor identity · Формула разности/суммы кубов

x³−y³=(x−y)(x²+xy+y²), x³+y³=(x+y)(x²−xy+y²).

Sistema ikkinchi factorni allaqachon berganda x±y darhol topiladi.

Misol: x³−y³=7 va x²+xy+y²=7 → x−y=1.

Bu emas: x³−y³=(x−y)³ emas.

💡 6958–6960 tipidagi savollar yadrosi.

Sum-of-squares criterion · Sum-of-squares criterion · Критерий суммы квадратов

Real sonlarda A²+B²=0 iff A=B=0.

Bitta nonlinear equation ikki linear/nonlinear conditionga ajralishi mumkin.

Misol: (x−2)²+(y+5)²=0 → (2,−5).

Bu emas: A²+B²=0 dan A=−B deyish noto‘g‘ri.

💡 Juft yuqori darajalarda ham nonnegative argument ishlaydi.

Cyclic sistema · Cyclic system · Циклическая система

Tenglamalar variablelar cyclic tartibda bir-biriga o‘tadigan sistema.

x→y→z→x struktura mavjud.

Misol: x²+4=4y, y²+4=4z, z²+4=4x.

Bu emas: Tasodifiy uch tenglamali sistema cyclic emas.

💡 Symmetriya, AM-GM yoki summation argumentlari ishlashi mumkin.

Geometrik kesishish · Geometric intersection · Геометрическое пересечение

Sistema yechimlari tegishli grafiklarning umumiy nuqtalari.

Line-circle 0,1 yoki 2 kesishish nuqtasiga ega bo‘lishi mumkin.

Misol: y=x+1 va x²+y²=25.

Bu emas: Har graphdagi alohida nuqta solution emas.

💡 Yechimlar sonini oldindan tasavvur qilishga yordam beradi.

Solution multiplicity vs ordered solutions · Ordered-solution count · Число упорядоченных решений

Bir xil unordered sonlar jufti symmetric sistemada ikki ordered solution berishi mumkin.

(a,b) va (b,a), a≠b, alohida yechimlar.

Misol: x+y=6,xy=8 → (2,4),(4,2).

Bu emas: (3,3) ikki marta sanalmaydi.

💡 Solution count savollarida muhim.

Independent verification · Independent verification · Независимая проверка

Candidate pairni original sistema va domain bilan mustaqil tekshirish.

Marked option yoki intermediate factorization proof o‘rnini bosmaydi.

Misol: Har pairni barcha equationsga substitution.

Bu emas: Faqat correct_optionga qarash verification emas.

💡 Bankdagi malformed/wrong records sabab canonical QA uchun majburiy.

Fundamental tushunchalar

All-equations verification

Ordered pair final yechim bo‘lishi uchun sistemaning har bir original tenglamasini qanoatlantirishi kerak.

$(a,b)\in S\iff E_1(a,b)=\cdots=E_n(a,b)=0$

Bitta tenglamaga mos kelish yetarli emas.

Substitution pipeline

Bir equationdan y=f(x) yoki x=g(y) oling, boshqa equationga qo‘ying, bir variable equationni yeching, so‘ng pairlarni tiklang.

$y=f(x)\Rightarrow G(x,f(x))=0$

Har x rootdan y qiymati alohida tiklanadi.

Elimination pipeline

Ikki equationdagi mos nonlinear hadlar koeffitsientlarini teng/opposite qilib qo‘shish yoki ayirish.

$E_1\pm kE_2$

x² yoki y² hadini yo‘qotish mumkin.

Graph interpretation

Har equation tekislikda egri chiziq, sistemaning yechimlari ularning kesishish nuqtalari.

0,1,2 yoki undan ko‘p real kesishish mumkin.

Sum-product coordinates

Symmetric sistemada x,y o‘rniga S=x+y va P=xy ishlatiladi.

$S=x+y,\quad P=xy$

x,y keyin t²−St+P=0 roots sifatida tiklanadi.

Square-sum invariant

x²+y² symmetric ifoda S,P orqali yoziladi.

$x^2+y^2=S^2-2P$

S yoki P ma’lum bo‘lsa ikkinchisini topish mumkin.

Difference invariant

x−y=D va x²−y²=K bo‘lsa D(x+y)=K orqali sum olinadi.

$x+y=K/D$

D≠0; D=0 alohida branch.

Cube-difference reduction

x³−y³ va x²+xy+y² berilsa factor identity x−y ni beradi.

$x^3-y^3=(x-y)(x^2+xy+y^2)$

Ikkinchi factor nonzero bo‘lsa division mumkin.

Cube-sum reduction

x³+y³ va x²−xy+y² berilsa x+y olinadi.

$x^3+y^3=(x+y)(x^2-xy+y^2)$

Factor zero branchni tekshiring.

Symmetric subtraction factor

F(x,y)=F(y,x) ko‘rinishidagi ikki equationni ayirish ko‘pincha (x−y)H(x,y)=0 beradi.

$(x-y)H(x,y)=0$

x=y va H=0 branchlarning ikkalasi ham yechiladi.

Common-sum factor

x²+xy=A va y²+xy=B bo‘lsa x(x+y)=A, y(x+y)=B.

$xs=A,\ ys=B,\ s=x+y$

s=0 imkonini alohida tekshiring.

Ratio invariant

x/y=r kabi relation berilsa x=ry bilan dimension kamayadi.

$x=ry$

y≠0.

Reciprocal sum invariant

x,y nonzero bo‘lsa 1/x+1/y=(x+y)/(xy)=S/P.

$\frac1x+\frac1y=\frac SP$

P≠0.

Zero-product branch tree

Har product=0 equation factor choices beradi; kombinatsiyalar branch tree orqali kesishadi.

$AB=0\Rightarrow A=0\text{ yoki }B=0$

Har branch boshqa equations bilan kesishadi.

Domain in systems

Kasrli sistemada barcha denominator restrictionlar barcha branchlar uchun global constraint.

$D=\{(x,y):Q_1\cdots Q_m\ne0\}$

Factor zero candidate denominatorni nol qilsa rad etiladi.

Sum of nonnegative terms

Real kvadratlar yoki juft darajalar yig‘indisi 0 bo‘lsa har term 0.

$A^2+B^2=0\Rightarrow A=B=0$

Termlar real-valued.

Count ordered solutions

Symmetric pair a≠b bo‘lsa permutationlar ikki ordered solution; a=b bo‘lsa bitta.

$(a,b),(b,a)$

Duplicate ordered pair qayta sanalmaydi.

Rectangle invariant model

Tomonlar a,b uchun perimeter 2S, area P, diagonal²=S²−2P.

$d^2=a^2+b^2=S^2-2P$

a,b>0.

Cyclic equality strategy

Cyclic systemda tenglamalarni qo‘shish, monotonicity yoki nonnegative differences orqali x=y=z ni majburlash mumkin.

$x\to y\to z\to x$

Specific coefficientsga qarab proof tanlanadi.

Candidate reconstruction

Invariant S,P topilgach x,y ni t²−St+P rootsidan ordered pairlarga aylantiring.

$t^2-St+P=0$

Discriminant real pair existence’ni tekshiradi.

Malformed-source gate

Savolda e’lon qilinmagan variable yoki javob formatiga mos kelmaydigan unknown paydo bo‘lsa canonical misolga kiritilmaydi.

Source context/OCR tekshiriladi.

Branch completeness QA

Factorizationdan chiqqan barcha algebraik branchlar ro‘yxat qilinmaguncha yechim tugamagan hisoblanadi.

Har branch → candidates → original-system check.

Formula kutubxonasi

System solution criterion

$$(a,b)\in S\iff E_i(a,b)=0\quad\forall i$$

Pair barcha equationsni qanoatlantirishi kerak.

Substitution

$$y=f(x),\quad G(x,y)=0\Rightarrow G(x,f(x))=0$$

Ikki variable sistemani bir variablega tushiradi.

Symmetric sum-product

$$S=x+y,\quad P=xy$$

Symmetric sistemaning asosiy invariantlari.

Reconstruction quadratic

$$t^2-St+P=0$$

x va y shu quadratic roots.

Square sum

$$x^2+y^2=S^2-2P$$

Square sumni sum/productga aylantiradi.

Difference square

$$(x-y)^2=S^2-4P$$

Difference magnitude symmetric invariant.

Difference of squares

$$x^2-y^2=(x-y)(x+y)$$

Difference + sumni bog‘laydi.

Cube difference

$$x^3-y^3=(x-y)(x^2+xy+y^2)$$

Cubic systemni linear differencega tushiradi.

Cube sum

$$x^3+y^3=(x+y)(x^2-xy+y^2)$$

Cubic systemni linear sumga tushiradi.

Quadratic symmetric factor

$$x^2+xy+y^2=S^2-P$$

Positive-style cubic factorni S,P bilan yozadi.

Alternating quadratic factor

$$x^2-xy+y^2=S^2-3P$$

Cube-sum factorni S,P bilan yozadi.

Reciprocal sum

$$\frac1x+\frac1y=\frac{x+y}{xy}=\frac SP$$

Reciprocal equationni invariantga aylantiradi.

Shart: x,y≠0

Ratio substitution

$$\frac xy=r\Rightarrow x=ry$$

Ratio system dimensionini kamaytiradi.

Shart: y≠0

Common-sum products

$$x^2+xy=xS,\quad y^2+xy=yS$$

Ikki equationsni xS,yS ga aylantiradi.

Shart: S=x+y

Zero-product

$$A_1A_2\cdots A_n=0\iff A_1=0\text{ yoki }\cdots\text{ yoki }A_n=0$$

Har factor alohida branch.

Sum of squares zero

$$A^2+B^2=0\iff A=0,\ B=0$$

Bitta equationni ikki conditionga ajratadi.

Shart: A,B real

Even powers zero

$$A^{2m}+B^{2n}=0\iff A=B=0$$

Juft darajalar nonnegative.

Shart: m,n musbat butun; real A,B

Rectangle perimeter

$$\mathcal P=2S$$

Tomonlar yig‘indisidan perimeter.

Shart: S=a+b

Rectangle diagonal

$$d^2=S^2-2P$$

Diagonal, sum va areani bog‘laydi.

Shart: P=ab

Sum from diagonal and area

$$S=\sqrt{d^2+2P}$$

Rectangle tomonlar yig‘indisini topadi.

Shart: a,b>0

Product from sum and square sum

$$P=\frac{S^2-(x^2+y^2)}2$$

Square sumdan productni chiqaradi.

Real reconstruction condition

$$\Delta_t=S^2-4P\ge0$$

S,P dan real x,y mavjudlik sharti.

Teoremalar va isbotlar

📐 Substitution ekvivalentligi teoremasi

Agar bir equation y=f(x) ga ekvivalent bo‘lsa, sistemani {y=f(x), G(x,y)=0} dan G(x,f(x))=0 ga tushirish yechimlarni saqlaydi; topilgan x lar uchun y=f(x) qayta tiklanadi.

Graph intersectionni algebraik projection qilib ko‘ramiz.

Isbotni ko'rsatish

Berilgan: Sistema y=f(x) va G(x,y)=0 ko‘rinishida.

Isbotlash kerak: Substitutiondan keyingi x roots bilan original ordered pairs orasidagi moslikni ko‘rsatish.

Agar (a,b) original solution bo‘lsa, b=f(a) va G(a,b)=0.
Demak G(a,f(a))=0, shuning uchun a reduced equation rootidir.
Aksincha a reduced equation rooti bo‘lsa, b=f(a) deb oling.
Shunda y=f(x) ham, G(x,y)=0 ham rost; pair original system solution.

Equivalence saqlangan substitution solution setni to‘liq beradi. ∎

📐 Symmetric sum-product reconstruction teoremasi

Agar x+y=S va xy=P bo‘lsa, x va y t²−St+P=0 tenglamaning ildizlaridir; aksincha shu quadraticning ikki ildizini ordered tarzda joylashtirish sistemani qanoatlantiradi.

Ikki noma’lum sonni ularning sum va producti aniqlaydi.

Isbotni ko'rsatish

Berilgan: x+y=S va xy=P.

Isbotlash kerak: x,y t²−St+P=0 roots ekanini ko‘rsatish.

(t−x)(t−y)=t²−(x+y)t+xy.
x+y=S va xy=P ni qo‘ying.
Natija t²−St+P.
Demak t=x va t=y shu quadratic roots.
Aksincha quadratic roots sum S va product P bo‘lgani uchun ordered placements original invariant sistemani qanoatlantiradi.

Sum-product invariantlardan variables Viyet orqali tiklanadi. ∎

📐 Symmetric subtraction branch teoremasi

F(x,y)=C va F(y,x)=C tenglamalarni ayirish factorization orqali (x−y)H(x,y)=0 bersa, barcha yechimlar x=y branch yoki H(x,y)=0 branchlarning birida yotadi.

Symmetriya diagonal va off-diagonal yechimlarni ajratadi.

Isbotni ko'rsatish

Berilgan: F(x,y)=C va F(y,x)=C; ayirish (x−y)H(x,y)=0 beradi.

Isbotlash kerak: Har solution ikki branchdan birida ekanini ko‘rsatish.

Original system solution uchun ikki tenglama ayirmasi 0.
Factorizationdan (x−y)H(x,y)=0.
Zero-productga ko‘ra x−y=0 yoki H(x,y)=0.
Shuning uchun barcha solutions x=y yoki H=0 branchlaridan birida.
Har branch original equations bilan yana tekshiriladi.

Symmetric systemda diagonal va off-diagonal branchlarning ikkalasi ham majburiy. ∎

📐 Zero-product branch completeness teoremasi

A_1...A_m=0 va B_1...B_n=0 sistema yechimlari A_i=0 va B_j=0 branch juftliklarining barcha mos kesishmalaridan iborat.

Product zero bo‘lsa kamida bitta factor zero.

Isbotni ko'rsatish

Berilgan: A_1...A_m=0 va B_1...B_n=0.

Isbotlash kerak: Solution set barcha factor-branch kesishmalarining birlashmasi ekanini ko‘rsatish.

Birinchi product zero bo‘lsa kamida bir A_i=0.
Ikkinchi product zero bo‘lsa kamida bir B_j=0.
Demak har solution kamida bir (A_i=0,B_j=0) branch kesishmasida yotadi.
Aksincha bir branch ikkala productni zero qilsa va domain restrictionsni buzmasa, original product equations rost.

Branch tree to‘liq ko‘rilsa zero-product sistema yechimlari yo‘qolmaydi. ∎

📐 All-equations verification teoremasi

Algebraik reductiondan chiqqan kandidat tuple final solution bo‘lishi uchun original sistemadagi har bir tenglamani va barcha domain restrictionlarni qanoatlantirishi zarur va yetarli.

System conditionsning kesishmasidir.

Isbotni ko'rsatish

Berilgan: Candidate tuple c va original system E_i=0 hamda domain D.

Isbotlash kerak: c final solution iff c∈D va barcha E_i(c)=0.

Zarurlik: solution ta’rifiga ko‘ra c barcha original equationsni qanoatlantiradi va expressions aniqlangan bo‘lishi kerak.
Yetarlilik: c domain ichida bo‘lib har bir original equationni rost qilsa, u system conditionsning umumiy kesishmasida.
Shuning uchun intermediate reductiondagi candidate status final verificationgacha yetarli emas.

Final solution faqat all-equations/domain checkdan o‘tgan candidate. ∎

Yechilgan misollar

oson Quyidagi sistemani yeching: $\begin{cases}x+y=6\\xy=8\end{cases}$

💡 Maslahat: S=6,P=8 dan reconstruction quadratic tuzing.

t²−6t+8=0.
(t−2)(t−4)=0, roots 2 va 4.
Ordered pairs: (2,4) va (4,2).

✅ Javob: (2,4),(4,2)

Nega bu usul ishlaydi: Symmetric systemda x,y quadratic rootsning ikki tartibli joylashuvi.

⚠️ (2,4) va (4,2) alohida ordered solutions.

oson $\begin{cases}x-y=3\\xy=10\end{cases}$ sistemani yeching.

💡 Maslahat: x=y+3 ni qo‘ying.

x=y+3.
y(y+3)=10 → y²+3y−10=0.
y=2 yoki −5.
Mos x=5 yoki −2.

✅ Javob: (5,2),(-2,-5)

Nega bu usul ishlaydi: Substitution systemni quadraticga tushiradi.

⚠️ Ikkinchi y rootdan keladigan pairni tashlamang.

oson $\begin{cases}x-y=3\\x^2-y^2=21\end{cases}$ sistemani yeching.

💡 Maslahat: x²−y²=(x−y)(x+y).

3(x+y)=21.
x+y=7.
x−y=3 va x+y=7 ni yeching: 2x=10.
x=5,y=2.

✅ Javob: (5,2)

Nega bu usul ishlaydi: Difference-of-squares nonlinear equationni linear sumga aylantiradi.

⚠️ x−y=0 bo‘lsa divisionga ehtiyot bo‘ling; bu misolda 3 nonzero.

oson $\begin{cases}x+y=6\\x^2+y^2=20\end{cases}$ sistemani yeching.

💡 Maslahat: x²+y²=S²−2P.

20=36−2P → P=8.
t²−6t+8=0 → t=2,4.
Ordered pairs (2,4),(4,2).

✅ Javob: (2,4),(4,2)

Nega bu usul ishlaydi: Square sum S,P invariantga tushadi.

⚠️ Faqat bitta permutationni yozmang.

ortacha $\begin{cases}x+y=2\\x^2+y^2-2xy=16\end{cases}$ sistemani yeching.

💡 Maslahat: Ikkinchi equation (x−y)²=16.

x−y=±4.
x+y=2 bilan x−y=4 → (3,−1).
x−y=−4 → (−1,3).

✅ Javob: (3,-1),(-1,3)

Nega bu usul ishlaydi: Square equation ikki sign branch beradi.

⚠️ Faqat +4 branchni olish bir solutionni yo‘qotadi.

oson $\begin{cases}y=x+6\\x^2+3=4y\end{cases}$ sistemani yeching.

💡 Maslahat: Birinchi equationni ikkinchisiga qo‘ying.

x²+3=4(x+6).
x²−4x−21=0.
x=7 yoki −3.
y=x+6 → y=13 yoki 3.

✅ Javob: (7,13),(-3,3)

Nega bu usul ishlaydi: Line equation variable’ni bevosita ifodalaydi.

⚠️ Har x rootga mos y ni alohida hisoblang.

ortacha $\begin{cases}y-3x=2\\x^2=2y+3\end{cases}$ sistemani yeching.

💡 Maslahat: y=3x+2.

x²=2(3x+2)+3=6x+7.
x²−6x−7=0 → x=7 yoki −1.
y=23 yoki −1.

✅ Javob: (7,23),(-1,-1)

Nega bu usul ishlaydi: Substitution bir quadratic hosil qiladi.

⚠️ Signni y=3x+2 da tekshiring.

ortacha $\begin{cases}x-2y=1\\3x+y^2=10\end{cases}$ sistemani yeching.

💡 Maslahat: x=1+2y.

3(1+2y)+y²=10.
y²+6y−7=0 → y=1 yoki −7.
x=3 yoki −13.

✅ Javob: (3,1),(-13,-7)

Nega bu usul ishlaydi: Linear relation quadratic equationni bitta variableda beradi.

⚠️ Bank 7019 dagi ikkala pair ham saqlanadi.

murakkab $\begin{cases}x^2+y=20\\y^2+x=20\end{cases}$ sistemani to‘liq yeching.

💡 Maslahat: Tenglamalarni ayirib factorlang.

Ayirish: x²−y²+y−x=(x−y)(x+y−1)=0.
Branch 1: x=y. x²+x=20 → x=4 yoki −5 → (4,4),(-5,-5).
Branch 2: x+y=1. y=1−x ni x²+y=20 ga qo‘ying.
x²−x−19=0 → x=(1±√77)/2; y=(1∓√77)/2.

✅ Javob: $(4,4),(-5,-5),\left(\frac{1+\sqrt{77}}2,\frac{1-\sqrt{77}}2\right),\left(\frac{1-\sqrt{77}}2,\frac{1+\sqrt{77}}2\right)$

Nega bu usul ishlaydi: Symmetric subtraction diagonal va off-diagonal branchlarni to‘liq beradi.

⚠️ Bank 6954 faqat diagonal ikki pairni bergan; off-diagonal branchlarni tashlash noto‘g‘ri.

murakkab $\begin{cases}x^2+2y=15\\y^2+2x=15\end{cases}$ sistemani yeching.

💡 Maslahat: Ayirishdan (x−y)(x+y−2)=0.

Branch x=y: x²+2x=15 → x=3,−5.
Branch x+y=2: y=2−x.
x²+4−2x=15 → x²−2x−11=0.
x=1±2√3; y=1∓2√3.

✅ Javob: $(3,3),(-5,-5),(1-2\sqrt3,1+2\sqrt3),(1+2\sqrt3,1-2\sqrt3)$

Nega bu usul ishlaydi: Symmetric factor branch method barcha yechimlarni beradi.

⚠️ Radical off-diagonal solutions ham real solutions.

ortacha $\begin{cases}x^3-y^3=7\\x^2+xy+y^2=7\end{cases}$ sistemani yeching.

💡 Maslahat: Cube difference factoridan x−y ni toping.

x³−y³=(x−y)(x²+xy+y²).
7=(x−y)·7 → x−y=1.
y=x−1 ni quadratic factorga qo‘ying: 3x²−3x+1=7.
x²−x−2=0 → x=2,−1; y=1,−2.

✅ Javob: (2,1),(-1,-2)

Nega bu usul ishlaydi: Matching cubic factor system degree’ni keskin kamaytiradi.

⚠️ x³−y³ ni (x−y)³ deb olmang.

ortacha $\begin{cases}x^3+y^3=28\\x^2-xy+y^2=7\end{cases}$ sistemani yeching.

💡 Maslahat: Cube sum factoridan x+y ni oling.

28=(x+y)·7 → x+y=4.
x²−xy+y²=S²−3P=7.
16−3P=7 → P=3.
t²−4t+3=0 → roots 1,3.

✅ Javob: (1,3),(3,1)

Nega bu usul ishlaydi: Cube-sum factor + S,P reconstruction.

⚠️ Ordered permutationsni ikkalasini yozing.

murakkab $\begin{cases}x^4+x^2y^2+y^4=91\\x^2+xy+y^2=13\end{cases}$ sistemani yeching.

💡 Maslahat: Birinchi expression ikki quadratic factor productidir.

x⁴+x²y²+y⁴=(x²+xy+y²)(x²−xy+y²).
Ikkinchi factor 91/13=7.
Ikki factorni ayiring: 2xy=6 → P=3.
x²+y²=10; S²=10+6=16 → S=±4.
S=4,P=3 → 1,3; S=−4,P=3 → −1,−3.

✅ Javob: (1,3),(3,1),(-1,-3),(-3,-1)

Nega bu usul ishlaydi: Higher-degree symmetric system elementary invariantsga tushadi.

⚠️ S=+4 branch bilan cheklanib qolmang; S²=16 ikki sign beradi.

ortacha $\begin{cases}x^2+xy=20\\y^2+xy=5\end{cases}$ sistemani yeching.

💡 Maslahat: s=x+y deb yozing.

x(x+y)=20 va y(x+y)=5.
s=x+y; xs=20, ys=5.
s²=25 → s=±5.
s=5 → (x,y)=(4,1); s=−5 → (−4,−1).

✅ Javob: (4,1),(-4,-1)

Nega bu usul ishlaydi: Common sum factor variablesni ratio bilan aniqlaydi.

⚠️ s²=25 dan faqat s=5 ni olmang.

ortacha $\begin{cases}x+y=7\\\frac1x+\frac1y=\frac7{12}\end{cases}$ sistemani yeching.

💡 Maslahat: Reciprocal sum S/P.

ODZ: x,y≠0.
S=7 va S/P=7/12 → P=12.
t²−7t+12=0 → t=3,4.
Ordered pairs (3,4),(4,3).

✅ Javob: (3,4),(4,3)

Nega bu usul ishlaydi: Reciprocal equation bevosita productni beradi.

⚠️ x=0 yoki y=0 candidate emas.

oson $\begin{cases}x+y=4\\\frac1x+\frac1y=1\end{cases}$ butun yechimlarni toping.

💡 Maslahat: S=4, S/P=1.

ODZ: x,y≠0.
P=4.
t²−4t+4=(t−2)².
x=y=2.

✅ Javob: (2,2)

Nega bu usul ishlaydi: Repeated quadratic root bitta ordered pair beradi.

⚠️ (2,2) ni ikki marta sanamang.

ortacha $\begin{cases}xy=6\\\frac1x-\frac1y=\frac16\end{cases}$ butun yechimlarni toping.

💡 Maslahat: Reciprocal difference numerator y−x.

ODZ: x,y≠0.
(y−x)/(xy)=1/6 va xy=6 → y−x=1.
y=x+1; x(x+1)=6.
x=2 yoki −3 → pairs (2,3),(-3,-2).

✅ Javob: (2,3),(-3,-2)

Nega bu usul ishlaydi: Product known bo‘lgani reciprocal difference linear differencega aylanadi.

⚠️ 1/x−1/y numerator y−x, x−y emas.

ortacha $\begin{cases}(x-5)(y-3)=0\\(x-8)(y+2)=0\end{cases}$ sistemani yeching.

💡 Maslahat: Har product uchun branch tree tuzing.

Birinchi: x=5 yoki y=3.
Ikkinchi: x=8 yoki y=−2.
x=5,x=8 inconsistent.
x=5,y=−2 → (5,−2).
y=3,x=8 → (8,3).
y=3,y=−2 inconsistent.

✅ Javob: (5,-2),(8,3)

Nega bu usul ishlaydi: Zero-product branchlarning barcha compatible kesishmalari olinadi.

⚠️ Branch kombinatsiyalaridan birini unutish yechim yo‘qotadi.

murakkab $\begin{cases}\frac{(x-5)^2}{y-3}=0\\\frac{(y-3)^2}{x-5}=0\end{cases}$ sistemani yeching.

💡 Maslahat: Zero rational expression + domainni birga ko‘ring.

Birinchi equation x=5 va y≠3 talab qiladi.
Ikkinchi equation y=3 va x≠5 talab qiladi.
Bu talablar bir vaqtda bajarilmaydi.

✅ Javob: $\varnothing$

Nega bu usul ishlaydi: Har equation numerator-zero va denominator-nonzero condition beradi; ularning kesishmasi bo‘sh.

⚠️ (5,3) ikkala denominatorni nol qiladi, solution emas.

oson (x-2)^2+(y+5)^2=0 tenglamani real sonlarda yeching.

💡 Maslahat: Har kvadrat nonnegative.

Yig‘indi 0 bo‘lishi uchun x−2=0 va y+5=0.
x=2,y=−5.

✅ Javob: (2,-5)

Nega bu usul ishlaydi: Sum-of-squares criterion bitta equationni ikki simultaneous conditionga aylantiradi.

⚠️ A=−B branchi yo‘q; har kvadrat alohida 0.

oson (3x-6)^4+(y-5)^4=0 tenglamani real sonlarda yeching.

💡 Maslahat: Juft darajalar nonnegative.

(3x−6)^4=0 va (y−5)^4=0.
3x−6=0 → x=2.
y−5=0 → y=5.

✅ Javob: (2,5)

Nega bu usul ishlaydi: Nonnegative even powers yig‘indisi 0 bo‘lsa har term 0.

⚠️ To‘rtinchi darajadan ± root branch izlash bu zero case’da keraksiz.

ortacha Diagonali $\sqrt{13}$, yuzi 6 bo‘lgan to‘g‘ri to‘rtburchak perimetrini toping.

💡 Maslahat: Tomonlar a,b; P=ab=6, d²=a²+b²=13.

S=a+b.
S²=a²+b²+2ab=13+12=25.
Tomonlar musbat, S=5.
Perimetr=2S=10.

✅ Javob: 10

Nega bu usul ishlaydi: Geometry sum-product invariant systemga aylanadi.

⚠️ S=±5 dan musbat tomonlar sabab S=5.

ortacha Diagonali $\sqrt{74}$, perimetri 24 bo‘lgan to‘g‘ri to‘rtburchak yuzini toping.

💡 Maslahat: S=12.

a+b=12.
74=a²+b²=S²−2P=144−2P.
2P=70 → P=35.
Area=ab=35.

✅ Javob: 35

Nega bu usul ishlaydi: Diagonal square identity productni beradi.

⚠️ Perimeter 24 bo‘lsa S=12, 24 emas.

murakkab $\begin{cases}x^2+y^2=100\\xy=-24\end{cases}$ sistemaning nechta real ordered yechimi bor?

💡 Maslahat: S²=x²+y²+2xy.

S²=100−48=52 → S=±2√13.
Har S uchun t²−St−24=0 discriminanti S²+96=148>0, ikkita turli real root beradi.
Har S ikki ordered placement beradi.
Jami 4 ordered solution.

✅ Javob: 4

Nega bu usul ishlaydi: Invariantlar explicit radicalsni hisoblamasdan solution countni beradi.

⚠️ S²=52 dan ikkala sign branchni saqlang.

oson $\begin{cases}x+y=10\\x^2+y^2=2xy+36\end{cases}$ sistemaning nechta real yechimi bor?

💡 Maslahat: Ikkinchi equation (x−y)²=36.

x−y=±6.
x+y=10 bilan pairs (8,2) va (2,8).
Jami 2 ordered solution.

✅ Javob: 2

Nega bu usul ishlaydi: Difference square ikki branch beradi.

⚠️ Pairlar bir-birining permutationi bo‘lsa ham ordered solutions sifatida alohida.

oson $\begin{cases}x+y=8\\(x-4)(y-3)=0\end{cases}$ sistemani qanoatlantiruvchi x lar yig‘indisini toping.

💡 Maslahat: Zero-productni ikki branchga ajrating.

x=4 branch: y=4 → x=4.
y=3 branch: x=5 → x=5.
x lar yig‘indisi 9.

✅ Javob: 9

Nega bu usul ishlaydi: Product equation branch tree bilan linear systemlarga ajraladi.

⚠️ Faqat x=4 branchni olmang.

murakkab $\begin{cases}x^3-y^3=26\\x-y=2\end{cases}$ bo‘lsa xy ni toping.

💡 Maslahat: Cube difference factorini ishlating.

26=2(x²+xy+y²) → x²+xy+y²=13.
(x−y)²=x²−2xy+y²=4.
Birinchidan ikkinchini ayiring: 3xy=9.
xy=3.

✅ Javob: 3

Nega bu usul ishlaydi: Explicit x,y ni topmasdan symmetric algebra productni beradi.

⚠️ x²+xy+y² va (x−y)² koeffitsientlarini to‘g‘ri ayiring.

murakkab $\begin{cases}x+y=3\\x^2+xy-y^2=5\end{cases}$ sistemani yeching.

💡 Maslahat: y=3−x qo‘ying.

x²+x(3−x)−(3−x)²=5.
Soddalashtirish: x²−9x+14=0.
x=2 yoki 7.
y=1 yoki −4.

✅ Javob: (2,1),(7,-4)

Nega bu usul ishlaydi: Known sum substitution nonlinear expressionni quadraticga tushiradi.

⚠️ Ikkinchi quadratic root 7 ni tashlamang.

murakkab $\begin{cases}\frac{x}{y}+x+y=7\\\frac{x}{y}(x+y)=12\end{cases}$ sistemani yeching.

💡 Maslahat: u=x/y, v=x+y deb oling.

ODZ: y≠0.
u+v=7, uv=12.
u,v roots 3 va 4.
u=3,v=4: x=3y,4y=4 → (3,1).
u=4,v=3: x=4y,5y=3 → (12/5,3/5).

✅ Javob: $(3,1),\left(\frac{12}{5},\frac35\right)$

Nega bu usul ishlaydi: Ikki murakkab invariantni yangi sistema sifatida yechish dimensionni saqlab strukturani soddalashtiradi.

⚠️ u va v roles almashganda yangi x,y pair paydo bo‘ladi.

murakkab $\begin{cases}x^2+4=4y\\y^2+4=4z\\z^2+4=4x\end{cases}$ sistemani real sonlarda yeching.

💡 Maslahat: Uch tenglamani qo‘shib perfect squares hosil qiling.

Tenglamalarni bir tomonga o‘tkazing va qo‘shing.
x²+y²+z²−4(x+y+z)+12=0.
Bu (x−2)²+(y−2)²+(z−2)²=0.
Har kvadrat 0: x=y=z=2.

✅ Javob: (2,2,2)

Nega bu usul ishlaydi: Cyclic system summation orqali nonnegative squares yig‘indisiga tushadi.

⚠️ Source optionlarda pair format berilgan bo‘lsa ham sistema uch variable; canonical javob tuple bo‘lishi kerak.

Umumiy xatolar

❌ Faqat bir tenglamani tekshirish.

System solution barcha equations kesishmasi.

✅ Har pairni barcha original equationsga qo‘ying.

6967 bank marked pair ikkinchi equationda yiqiladi.

❌ Symmetric systemda x=y branch bilan to‘xtash.

Ayirishdan H(x,y)=0 off-diagonal branch ham chiqishi mumkin.

✅ (x−y)H=0 ning ikkala branchini yeching.

6954 da ikki off-diagonal real solution bor.

❌ Ordered pair tartibini e’tiborsiz qoldirish.

(a,b) va (b,a) different assignments.

✅ Har pairni x,y rollari bilan yozing.

(2,4),(4,2).

❌ Repeated rootni ikki ordered solution deb sanash.

(a,a) permutation bilan o‘zgarmaydi.

✅ Duplicate pairni bir marta sanang.

(2,2).

❌ Substitutiondan keyingi x rootga mos y ni tiklamaslik.

Solution scalar emas, pair.

✅ Har x uchun original relationdan y ni toping.

y=x+6.

❌ Elimination qadamida equationni shartsiz bo‘lish.

Bo‘luvchi zero branch yo‘qolishi mumkin.

✅ Factor branchni alohida saqlang.

(x−y)H=0.

❌ Zero-productda bitta factorni tanlash.

Har factor zero bo‘lishi mumkin.

✅ Branch tree tuzing.

(x−5)(y−3)=0.

❌ Kasrli sistemada denominator restrictionni unutish.

Zero denominator pair original systemda aniqlanmagan.

✅ Global domainni oldindan yozing.

7033 da (5,3) solution emas.

❌ 1/x+1/y ni 1/(x+y) deb olish.

Kasrlar common denominator xy bilan qo‘shiladi.

✅ (x+y)/(xy)=S/P.

7040–7043.

❌ x³−y³ ni (x−y)³ deb olish.

Mixed terms mavjud.

✅ (x−y)(x²+xy+y²).

6958.

❌ x³+y³ ni (x+y)³ deb olish.

Mixed terms mavjud.

✅ (x+y)(x²−xy+y²).

6960.

❌ S²=value dan faqat musbat S olish.

Real S ikkala signga ega bo‘lishi mumkin.

✅ S=±sqrt(value) branchlarini tekshiring.

6961 tipida ± pairs.

❌ Sum-of-squares zero ni cancellation deb ko‘rish.

Real terms nonnegative.

✅ Har termni zero qiling.

(x−2)²+(y+5)²=0.

❌ Solution countda permutationsni unutish.

Symmetric system a≠b bo‘lsa ikki ordered pair beradi.

✅ Reconstructiondan keyin ordered placementsni yozing.

7000.

❌ Geometriyada perimeter’ni a+b deb olish.

Rectangle perimeter 2(a+b).

✅ Avval S=a+b, keyin 2S.

6987.

❌ 3-variable sistemaga pair formatdagi source javobni canonical deb olish.

Tuple dimension savol variable soniga mos bo‘lishi kerak.

✅ Original unknownlarni sanang va full tuple yozing.

6994 canonical (2,2,2).

❌ Savolda e’lon qilinmagan variable’ni taxmin bilan almashtirish.

Source/OCR corruption bo‘lishi mumkin.

✅ Malformed recordni canonical exampledan chiqarib qo‘ying.

6968 da z paydo bo‘ladi.

❌ correct_option ni verification o‘rnida ishlatish.

Bankda missing branch va inconsistent systems bor.

✅ Independent substitution/factorization bilan tekshiring.

6954,6967,6969,6974,6978.

Noto'g'ri tasavvurlar

Nochiziqli sistema faqat quadratic equationsdan iborat.

Rational, cubic, product, reciprocal va cyclic equations ham nonlinear system bo‘lishi mumkin.

Sistema yechimi bitta x qiymat.

Ikki variable sistema yechimi ordered pair, ko‘p variable sistema ordered tuple.

Symmetric sistemada x va y albatta teng.

Symmetriya x=y branchni beradi, lekin off-diagonal pairs ham mavjud bo‘lishi mumkin.

Substitution va elimination faqat linear sistemalar uchun.

Ikkala usul nonlinear sistemalarda ham qo‘llanadi.

Factorization branchlari orasidan eng qulay bittasini yechish yetarli.

Barcha algebraik branchlar ko‘rilishi kerak.

S=x+y va P=xy faqat Viyet mavzusiga tegishli.

Ular symmetric nonlinear systems uchun tabiiy koordinatalardir.

Kasrli sistema polynomialga aylangach domain restrictions kerak emas.

Original denominator zero nuqtalar doim excluded qoladi.

Grafik usul va algebraik usul turli yechim beradi.

To‘g‘ri qo‘llansa ikkalasi bir xil intersection pointsni beradi; grafik approximatsiya bo‘lishi mumkin.

Cyclic sistemada x=y=z ni taxmin qilishning o‘zi proof.

Equal tuple candidate; uniqueness uchun summation, monotonicity yoki boshqa argument kerak.

Marked test answer canonical matematik haqiqat.

Canonical content original equationsni mustaqil tekshirishga tayanadi.

Amaliy qo'llanilishi

Analitik geometriya

Line, circle, parabola va ellipse kesishish nuqtalari nonlinear systems orqali topiladi.

Geometriya

Rectangle side lengths diagonal, area va perimeter shartlaridan sum-product system sifatida tiklanadi.

Fizika

Bir nechta nonlinear constraint, masalan energiya va momentum tipidagi tenglamalar simultaneous system beradi.

Iqtisod va optimallashtirish

Nonlinear supply/demand yoki constraint equations bir nechta equilibrium nuqtalar berishi mumkin.

Computer algebra

Symbolic solvers substitution, Gröbner-type elimination va factor branchlarni avtomatlashtiradi, lekin verification baribir zarur.

Contest matematika

Symmetric invariants va cubic identities uzun eliminationni qisqa strukturaviy yechimga aylantiradi.

Model fitting

Ikki yoki undan ko‘p nonlinear relation parametr yoki state variablesni birgalikda aniqlaydi.

Test-bank QA

Ordered-pair substitution, variable-count va branch-completeness checks source extraction xatolarini ushlaydi.

Nochiziqli sistema: structure → branches → ordered solutions

Nochiziqli sistemada strukturani tanishdan final verified ordered pair/tuplegacha bo‘lgan pipeline: reduction, branch completeness, invariant reconstruction va all-equations QA.

Xulosa

Cheat sheet: 1) Yechim ordered pair/tuple; 2) substitution bir variable’ni boshqasiga ifodalaydi; 3) elimination mos hadlarni yo‘qotadi; 4) x+y=S, xy=P bo‘lsa x,y — t²−St+P=0 roots; 5) x²+y²=S²−2P; 6) x²−y²=(x−y)(x+y); 7) x³−y³=(x−y)(x²+xy+y²); 8) symmetric subtraction ko‘pincha (x−y)F(x,y)=0 branch beradi; 9) zero-productda barcha factor branchlar saqlanadi; 10) kasrli sistemada denominator restrictions oldindan yoziladi; 11) har ordered pair barcha original equationsda tekshiriladi.

Keyingi “Ko‘phadlar” va “Ratsional tengsizliklar” mavzularida factorization va sign analysis chuqurlashadi. “Parametrli tenglama va tengsizliklar”da esa solution branchlar soni parametrga qarab o‘zgaradi. Kvadrat funksiya mavzusida nonlinear systemlar grafik kesishish nuqtalari sifatida yana ko‘riladi.

Bog'liq mavzular

Oldin bilishingiz kerak: Kvadrat tenglama va uning ildizlari, Viyet teoremasi, Ratsional tenglamalar, Qisqa ko'paytirish formulalari

Bog'liq mavzular: Kvadrat funksiya

Keyingi mavzular: Parametrli tenglama va tengsizliklar, Ratsional tengsizliklar, Ko'phadlar

Manbalar

OpenStax Intermediate Algebra 2e — 11.5 Solve Systems of Nonlinear Equations — Graphing, substitution, elimination, applications and explicit final ordered-pair checking.
OpenStax Precalculus 2e — 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables — Definition of nonlinear systems, substitution, elimination and intersection-count interpretation.
Khan Academy — Quadratic systems: a line and a circle — Worked quadratic system interpreted as graph intersections.
Khan Academy — Quadratic system with no solutions — Algebraic and graphical no-real-intersection case.
OpenStax Intermediate Algebra — 11.5 Solve Systems of Nonlinear Equations — Substitution/elimination workflow and checking each ordered pair in both original equations.
OpenStax Elementary Algebra 2e — 5.3 Solve Systems of Equations by Elimination — Addition-property foundation for elimination, extended in this lesson to matching nonlinear terms.

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