If `x^(2)-7y=8, x-y=1`, and `y gt 0`, what is the value of y ?

Options

A. `1.64`
B. `4.78`
C. `5.64`
D. `6.14`

Answer

D

Explanation

Since you want the value of y, solve the equation `x - y=1` for x in terms of y. Then substitute for x the expression containing y that equals x in the equation `x^(2)-7y=8`. Adding y to both sides of the equation `x - y=1`, you have `x=y+1`. Now substitute `y+1` for the equation `x^(2)-7y=8`. You have `(y+1)^(2)-7y=8`. Solve this equation for y.
`(y+1)^(2)-7y=8`
`y^(2)+2y+1-7y=8`
`y^(2)-5y-7=0`
The general solution to the quadratic equation `ax^(2)+bx+c = 0` is
`x = (-b pm sqrt(b^(2)-4ac))/(2a)`
Here, you have the equation `y^(2)-5y-7=0` in the variable y. This equation is also a quadratic equation where the right side is 0. You have quadratic equation `ay^(2)+by+c=0`. where `a = 1, b =-5`, and c = 7. The solution to this equation is `y=(-(-5)pm sqrt((-5)^(2)-4(1)(-7)))/(2(1))=(5pm sqrt(25-(-28)))/(2)=`
`(5pm sqrt(53))/(2)`
Now, the question stem says that `y gt 0`. Since `sqrt(53)gt sqrt(49)=7, 5-sqrt(53)` is negative, so `(5-sqrt(23))/(2)` calculator, `(5+sqrt(23))/(2)~~6.14`. Thus, `y ~~ 6.14`.