From the first equation,
x+5 = 3(y+5)
x = 3y + 15 - 5
Now substitue x in the second equation with (3y +15 - 5).
x-5 = 7(y-5)
(3y+15-5) - 5 = 7(y-5)
3y +5 = 7y - 35
-4y = - 40
y = 10

Since y is 10, and x is (3y +15 - 5),
x = 30 + 15 - 5 = 40

5 0

2 years ago

What are the solutions to the quadratic equation -2x^2 + 6x + 3 = 0?

mario62 [17]

For this, we will be using the quadratic formula, which is $x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$ , with a=x^2 coefficient, b=x coefficient, and c = constant. Our equation will look like this: $x=\frac{-6+/-\sqrt{6^2-4*(-2)*3}}{2*(-2)}$

Firstly, solve the multiplications and the exponents: $x=\frac{-6+/-\sqrt{36+24}}{-4}$

Next, do the addition: $x=\frac{-6+/-\sqrt{60}}{-4}$

Next, your equation will be split into two: $x=\frac{-6+\sqrt{60}}{-4},\frac{-6-\sqrt{60}}{-4}$ . Solve them separately, and your answer will be $x=-0.436,3.436$