Answer:
No real solution
Step-by-step explanation:
Given
[tex]y = x^2 + 12x + 30[/tex]
[tex]8x - y = 10[/tex]
Required
Find x and y
Make y the subject of formula in [tex]8x - y = 10[/tex]
[tex]8x - y + y = 10 + y[/tex]
[tex]8x = 10 + y[/tex]
Subtract 10 from both sides
[tex]8x - 10 = 10 - 10 + y[/tex]
[tex]8x - 10 = y[/tex]
Substitute 8x - 10 for y in [tex]y = x^2 + 12x + 30[/tex]
[tex]8x - 10 = x^2 + 12x + 30[/tex]
Subtract (8x - 10) from both sides
[tex]8x - 10 - (8x - 10) = x^2 + 12x + 30 - (8x - 10)[/tex]
[tex]0 = x^2 + 12x + 30 - (8x - 10)[/tex]
[tex]0 = x^2 + 12x + 30 - 8x + 10[/tex]
Collect like terms
[tex]0 = x^2 + 12x -8x + 30+ 10[/tex]
[tex]0 = x^2 + 4x + 40[/tex]
[tex]x^2 + 4x + 40 = 0[/tex]
At this point, we have a quadratic equation;
First, we'll check if the quadratic equation has real solution using:
[tex]d = b^2 - 4ac[/tex]
Where a = 1; b = 4 and c= 40
[tex]d = 4^2 - 4 * 1 * 40[/tex]
[tex]d = 16 - 160[/tex]
[tex]d = -144[/tex]
Because the value of d is negative, then we can conclude that the solution has no real solution
