Question

How do you solve 5 = y - x and `4x^2 = -17x + y + 4`?

Answer 1

I will let you finish the calculation

Explanation:

Given:

`5=y-x" ".......................Equation(1)`
`4x^2=-17x+y+4" "......Equation(2)`

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Consider equation(1)

Add `x` to both sides.

`5+x=y-x+x`
`5+x=y" "......................Equation(1_a)`

Using `Equation(1_a)` substitute for `color(red)(y)` in `Equation(2)`

`4x^2=-17x+color(red)(y)+4" "->" "4x^2=-17x+color(red)(5+x)+4`

`" "->" "4x^2=-16x+9`
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Now we have a quadratic equation

`4x^2=-16x+9" "->" "4x^2+16x-9=0`

Use the formula to solve this:

Standard form `-> y-ax^2+bx+c` where `x=(-b+-sqrt(b^2-4ac))/(2a)`

I will let you do the next bit

You should get `x=-2+-5/2`

By substitution you can then find the value of `y`

Answer 2

The parabola and the straight line meet at `( -9/2, 1/2) and (1/2, 11/2)`
Illustrative Socratic graphs are inserted.

Explanation:

The first equation is of the form

y = a quadratic in x, and so, represents a parabola.

The line y = x+5 cuts the parabola, when

#4x^2+16x-9=0, giving x = -9/2 and 1/2.

Correspondingly,

y = x + 5 = 1/2 and 11/2.

So, the common points are `(1/2, 11/2) and (-9/2, 1/2)`.

graph{(y-4x^2-17x+4)(y-x-5)=0 [-50, 50, -25, 25]}

graph{(y-4x^2-17x+4)(y-x-5)=0 [-20, 20, -10, 10]}