How do you solve $y = x^{2}$ $y=x^2$ and $y = 0.5 x + 5$ $y= 0.5x + 5$ using substitution?

Question

2755 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-02T14:59:13

Please see below.

Just substitute value of $y$ $y$ from equation $y = 0.5 x + 5$ $y=0.5x+5$ in $y = x^{2}$ $y=x^2$

and we get $0.5 x + 5 = x^{2}$ $0.5x+5=x^2$ and multiplying by $2$ $2$ , we get

$x + 10 = 2 x^{2}$ $x+10=2x^2$ or $2 x^{2} - x - 10 = 0$ $2x^2-x-10=0$

i.e. $2 x^{2} - 5 x + 4 x - 10 = 0$ $2x^2-5x+4x-10=0$

or $x (2 x - 5) + 2 (2 x - 5) = 0$ $x(2x-5)+2(2x-5)=0$

or $(x + 2) (2 x - 5) = 0$ $(x+2)(2x-5)=0$

i.e. either $x + 2 = 0$ $x+2=0$ i.e. $x = - 2$ $x=-2$

or $2 x - 5 = 0$ $2x-5=0$ i.e. $x = \frac{5}{2} = 2.5$ $x=5/2=2.5$

Putting these values of $x$ $x$ in $y = 0.5 x + 5$ $y=0.5x+5$ , we get

$y = 4$ $y=4$ or $y = 6.25$ $y=6.25$

How do you solve y=x2y=x^2 and y=0.5x+5y= 0.5x + 5 using substitution?