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Zorluk: OrtaSimultaneous Linear and Quadratic Equations

If (x,y)(x, y) satisfies the simultaneous equations x+2y=5x + 2y = 5 and x2+y2=10x^2 + y^2 = 10, what is the positive value of xx?

Cevap: 3

Cevap

The positive value of xx is 3.
Isolating xx in the linear equation gives x=5−2yx = 5 - 2y. Substituting this expression into x2+y2=10x^2 + y^2 = 10 yields (5−2y)2+y2=10(5 - 2y)^2 + y^2 = 10. Expanding gives 25−20y+4y2+y2=10  ⟹  5y2−20y+15=025 - 20y + 4y^2 + y^2 = 10 \implies 5y^2 - 20y + 15 = 0. Dividing all terms by 55 produces y2−4y+3=0y^2 - 4y + 3 = 0, which factors as (y−1)(y−3)=0(y - 1)(y - 3) = 0, so y=1y = 1 or y=3y = 3. Substituting these into x=5−2yx = 5 - 2y gives x=3x = 3 when y=1y = 1 and x=−1x = -1 when y=3y = 3. The positive value of xx is 3.

Adım Adım Çözüm

1
Express xx from the linear equation
x=5−2yx = 5 - 2y
Isolating xx allows substitution into the quadratic equation.
2
Substitute into the quadratic equation
(5−2y)2+y2=10(5 - 2y)^2 + y^2 = 10
Eliminates variable xx to create a single-variable equation in yy.
3
Expand and simplify
5y2−20y+15=0  ⟹  y2−4y+3=05y^2 - 20y + 15 = 0 \implies y^2 - 4y + 3 = 0
Transforms the equation into standard quadratic form for easy factorization.
4
Solve for yy
y=1 or y=3y = 1 \text{ or } y = 3
Factoring (y−1)(y−3)=0(y - 1)(y - 3) = 0 yields the two possible values for yy.
5
Determine corresponding xx values and select the positive one
x=3x = 3 (from y=1y = 1)
Evaluating x=5−2yx = 5 - 2y gives x=3x = 3 and x=−1x = -1; the positive result requested is 3.

Anahtar Kavram

Solving simultaneous linear and quadratic equations by substitution
Tahmini Süre:1m 30s
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Konular

Simultaneous Linear and Quadratic Equations

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