Thread: Help, please

I have one question left on my assignment, but it's confusing me, as algebra is not my strongest subject!!

If somebody could figure this out for me, it would be greatly appreciated!!

"Solve the following system algebraically. Show your work"

3/4x-2/5y=2

1/2x-3/5y=-2

Originally Posted by star4822

I have one question left on my assignment, but it's confusing me, as algebra is not my strongest subject!!

If somebody could figure this out for me, it would be greatly appreciated!!

"Solve the following system algebraically. Show your work"

3/4x-2/5y=2

1/2x-3/5y=-2

please clarify, is it:

(3/4)x-(2/5)y=2

(1/2)x-(3/5)y=-2

or

3/(4x)-2/(5y)=2

1/(2x) - 3/(5y)=-2

Sorry, it's meant to be:

(3/4)x-(2/5)y=2

(1/2)x-(3/5)y=-2

Originally Posted by star4822

Sorry, it's meant to be:

(3/4)x-(2/5)y=2

(1/2)x-(3/5)y=-2

ok, here goes. we will do the elimination method, rather than the substitution method

(3/4)x-(2/5)y=2 ..................................(1)

(1/2)x-(3/5)y=-2 .................................(2)

x - (8/15)y = 8/3 .................................(3) = (1) * (4/3)
x - (6/5)y = -4 ....................................(4) = (2) * 2

now subtract the (4) equation from the (3) equation, we get:

(2/3)y = 20/3 .....................................(5) = (3) - (4)
=> y = 10 ..........................................(6) = (5) * 3/2

but from equation (4), x - (6/5)y = -4
=> x - (6/5)(10) = -4
=> x = 60/5 - 4 = 8

so we get x = 8, y = 10