4pq(3-q)+2p(q-3)-7q(p-2pq)-2p(5q^2 +3)
the final result contains how many terms?

x=5.6

y=-5.6

Step-by-step explanation:

3x - 5y = 14 Equation 1

– 2x + 2y = 0 Equation 2

Simultaneous equation can be solved either through elimination method or substitution method. But we use elimination method for this question

Multiply equation 1 by -2 (the coefficient of x in equation 2) and multiply equation 2 with 3 (the coefficient of x in equation 1), so that x will have the same coefficient in the new equations, and easy to eliminate

-2(3x - 5y = 14)

-6x+10y=-28 Equation 3

3(– 2x + 2y = 0)

-6x+6y=0 Equation 4

Subtract equation 4 from 3 to eliminate x

-6x+10y=-28

-6x+6y=0

-6x-(-6x)=-6x+6x=0

10y-6y=5y

-28-0=-28

5y=-28

y=-28/5

y=-5.6

Substitute for y in equation 2

– 2x + 2y = 0

– 2x + 2(-5.6) = 0

-2x-11.2=0

-2x=11.2

x=-11.2/2

x=5.6