Question

Solve the following linear quadratic system of equations algebraically.

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

Answer 1

`x=1`

`y=2`

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`