# solve the systems pf equations alegbraically: y+4=3x y=x^2+5x-39

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y+ 4 = 3x...........(1)

y= x^2 + 5x -39 .............(2)

We will use the substitution method to solve.

We will substitute y into equation (1).

==> x^2 + 5x -39 + 4 = 3x

Now we will combine all terms.

==> x^2 + 2x - 35 = 0

Now we will factor.

==> (x+7) (x-5) = 0

==> x1= -7 ==> y1= 3x-4 = 3*-7 -4= -25

==> x2= 5 ==> y2 = 3*5 -4 = 11

Then we have two sets of solution for the system.

**The solution is the pair ( -7, -25) and ( 5, 11)**

The equations to be solved are

y + 4 = 3x ...(1)

y = x^2 + 5x - 39 ...(2)

(1)

=> y = 3x - 4

substitute in (2)

y = x^2 + 5x - 39

=> 3x - 4 = x^2 + 5x - 39

=> x^2 + 2x - 35 = 0

=> x^2 + 7x - 5x - 35 = 0

=> x(x + 7) - 5(x + 7) = 0

=> (x + 7)(x - 5) = 0

=> x = -7 and x = 5

For x = -7, y = -7*3 - 4 = -25

For x = 5, y = 3*5 - 4 = 11

**The solution of the equation is (-7 , -25) and (5 , 11)**