• Mar 27th 2007, 11:12 AM
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
• Mar 27th 2007, 11:16 AM
Jhevon
Quote:

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

(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
• Mar 27th 2007, 11:19 AM
star4822
Sorry, it's meant to be:

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

(1/2)x-(3/5)y=-2
• Mar 27th 2007, 11:50 AM
Jhevon
Quote:

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