How do you solve $y = x^{2}$ $y=x^2$ and $y = \frac{1}{2} x + 5$ $y=1/2x+5$ using substitution?

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How do you solve $y = x^{2}$ $y=x^2$ and $y = \frac{1}{2} x + 5$ $y=1/2x+5$ using substitution?

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Answer 1 · 2017-06-30T11:03:51

Solution: $(x = \frac{5}{2}, y = \frac{25}{4}) or (x = - 2, y = 4)$ $( x= 5/2 , y= 25/4) or (x= -2 , y=4)$

Explanation:

$y = x^{2} (1) and y = \frac{1}{2} x + 5 (2)$ $y= x^2 (1) and y =1/2 x +5 (2)$ . Substituting $y = \frac{1}{2} x + 5$ $y =1/2 x +5$ in equation (1) we get,

$x^{2} = \frac{1}{2} x + 5 or 2 x^{2} = x + 10 or 2 x^{2} - x - 10 = 0$ $x^2 =1/2 x +5 or 2x^2 =x+10 or 2x^2 -x -10 = 0$ or

$2 x^{2} - 5 x + 4 x - 10 = 0 or x (2 x - 5) + 2 (2 x - 5) = 0$ $2x^2 -5x +4x -10 = 0 or x(2x-5) + 2(2x-5) =0$ or

$(2 x - 5) (x + 2) = 0$ $(2x-5)(x+2) = 0$ . Either $2x-5 = 0 or 2x= 5 :. x=5/2$ or

$x+2=0 :. x=-2 :. x =5/2 or x= -2$

When $x=5/2 ; y= 1/2*5/2 +5 = 5/4+5 =25/4$ ;

When $x=-2 ; y= 1/2*(-2) +5 = -1+5 =4$ ;

Solution: $( x= 5/2 , y= 25/4) or (x= -2 , y=4)$ [Ans]

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How do you solve $y = x^{2}$ $y=x^2$ and $y = \frac{1}{2} x + 5$ $y=1/2x+5$ using substitution?

1 Answer

Explanation:

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How do you solve $y = x^{2}$ $y=x^2$ and $y = \frac{1}{2} x + 5$ $y=1/2x+5$ using substitution?