Question

CREATE A SYSTEM OF EQUATIONS AND FIND THE SOLUTION TO 5y-40x=10 y+6x=0

Answer 1

x = -1/7, y = 6/7

System:
`\left \{ {{5y-40x=10} \atop {y+6x=0}} \right.`

Explanation:

The system of equations would be
`\left \{ {{5y-40x=10} \atop {y+6x=0}} \right.`.

To find out what x and y are, choose one of the equation and solve for y. Then, substitute the result for y into the other equation.

5y - 40x = 10 (Add 40x to both sides)

5y = 40x + 10 (Divide both sides by 5)

y = 1/5(40x + 10) (Multiply 40x + 10 by 1/5)

y = 8x + 2 (Substitute this for y in the other equation.)

8x + 2 + 6x = 0 (Add 8x to 6x)

14x + 2 = 0 (Subtract 2 from both sides)

14x = -2 (Divide both sides by 14)

x = -1/7 (Substitute this for x in the y = 8x + 2)

y = 8(-1/7) + 2 (Multiply -1/7 by 8)

y = -8/7 + 2 (Convert 2 into 14/7; show as a single fraction)

y = (-8 + 14)/7 (Add -8 and 14)

y = 6/7 (The system is solved)

Answer 2

x = -1/7

y = 6/7

Explanation:

factor a 5 out of the first equation to work with smaller numbers:

y - 8x = 2

let y = -6x (derived from the second equation)

substitute: -6x - 8x = 2

-14x = 2

x = -2/14 or -1/7

find 'y': y - 8(-1/7) = 2

y + 8/7 = 14/7

y = 6/7