Question

Can I get some help solving this equation by using the substitution method? Thanks~

Answer 1

See a solution step process below;

Explanation:

Given;

`2y - x = 5 - - - eqn1`

`x^2 + y^2 - 25 = 0 - - - eqn2`

By Substitution Method!

From `eqn1`

`2y - x = 5`

`2y = 5 + x`

`y = (5 + x)/2 - - - eqn3`

Substitute the `y` into `eqn2`

`x^2 + y^2 - 25 = 0`

`x^2 + color(blue)(((5 + x)/2)^2) - 25 = 0`

`x^2 + ((5 + x)^2)/2^2 - 25 = 0`

`x^2 + (5 + x)^2/4 - 25 = 0`

Multiply through by `4`

`4(x^2) + 4((5 + x)^2/4) - 4(25) = 4(0)`

`4x^2 + cancel4((5 + x)^2/cancel4) - 100 = 0`

`4x^2 + (5 + x)^2 - 100 = 0`

`4x^2 + (5 + x) (5 + x) - 100 = 0`

`4x^2 + 25 + 5x + 5x + x^2 - 100 = 0`

`4x^2 + 25 + 10x + x^2 - 100 = 0`

`4x^2 + x^2 + 10x + 25 - 100 = 0`

`5x^2 + 10x - 75 = 0`

Divide through by `5`

`(5x^2)/5 + (10x)/5 - 75/5 = 0/5`

`(cancel5x^2)/cancel5 + (cancel10^2x)/cancel5 - cancel75^15/cancel5 = 0/5`

`x^2 + 2x - 15 = 0 -> "Quadratic Equation"`

Using Factorization Method!

Factors are; `+5 and -3`

Reason? `rArr +5 xx -3 = color(red)(-15) and +5 - 3 = color(red)(+2)`

Hence;

`x^2 + 2x - 15 = 0`

`x^2 - 3x + 5x - 15 = 0`

`(x^2 - 3x) (+ 5x - 15) = 0`

`x(x - 3) +5 (x - 3) = 0`

`(x - 3)(x + 5) = 0`

`x - 3 = 0 or x + 5 = 0`

`x = 3 or x = -5`

Substitute the value of `x` into `eqn3`

`y = (5 + x)/2`

When `x = 3`

`y = (5 + 3)/2`

`y = 8/2`

`y = 4`

When `x = -5`

`y = (5 + (-5))/2`

`y = (5 - 5)/2`

`y = 0/2`

`y = 0`

Therefore;

`x = -5, y = 0 and x = 3, y = 4`

Hope this helps!

Answer 2

`(x, y) = (-5, 0) and (3, 4)`

Explanation:

`2y - x = 5` => eq-1
`x^2+y^2-25=0` => eq-2
Find x in terms of y in eq-1:
`x=2y-5` => eq-3
Substitute for x in eq-2, solve the quadratic for y:
`(2y-5)^2+y^2=25`
`4y^2-20y+25+y^2=25`
`5y^2-20y=0`
`5y(y-4)=0`
`y=0, 4`
Plug in eq-3, solve for x:
`x=-5, 3`
Thus the solutions are:
`(x, y) = (-5, 0) and (3, 4)`