Question

Solve the following linear quadratic system of equations algebraically. y=^2+3x-2 y+3=5x

Answer 1

Answer:

$x=1$

$y=2$

Step-by-step explanation:

$y=x^2+3x-2$ , $y+3=5x$

Replace all occurrences of $y$ in $y+3=5x$ with $x^2+3x-2.$

$(x^2+3x-2)+3=5x$

$y=x^2+3x-2$

Add $-2$ and 3.

$x^2+3x+1=5x$

$y=x^2+3x-2$

Subtract 5x from both sides of the equation.

$x^2+3x+1-5x=0$

$y=x^2+3x-2$

Subtract 5x from 3x.

$x^2-2x+1=0$

$y=x^2+3x-2$

Rewrite 1 as $1^2$.

$x^2-2x+1^2=0$

$y=x^2+3x-2$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2x=2$ · $x$ · $1$

$y=x^2+3x-2$

Rewrite the polynomial.

$x^2-2$ · $x$ · $1$ $+$ $1^2=0$

$y=x^2+3x-2$

Factor using the perfect square

trinomial rule $a^2-2ab+b^2=(a-b)^2,$

where a = x and b = 1.

$(x-1)^2=0$

$y=x^2+3x-2$

Set the $x-1$ equal to 0.

$x-1=0$

$y=x^2+3x-2$

Add 1 to both sides of the equation.

$x=1$

$y=x^2+3x-2$

Replace all occurrences of $x$ in

$y=x^2+3x-2$ with 1.

$y=(1)^2+3(1)-2$

$x=1$

$y=2$