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

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#

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]}