2006-06-22 00:44:41 · 7 answers · asked by marcylina m 1 in Science & Mathematics ➔ Mathematics
Solve the first one for x: x = 3
Solve the second one for y:
7x - 5y = 8
7(3) - 5y = 8
21 - 5y = 8
-5y = -13
y = 13/5
Check it:
3-3 = 0 ---> OK
7(3) - 5(13/5) = 21 - 13 = 8 --->OK
Done. Hope this helps!
2006-06-22 00:50:46 · answer #1 · answered by anonymous 7 · 0⤊ 0⤋
x=3
7*3-8=13
5y=13
y=13/5
2006-06-22 08:18:52 · answer #2 · answered by ashok reddy 2 · 0⤊ 0⤋
is this one problem? if it is move the -3 over, it will be x=3. then plug the 3 into the next problem. 7(3)-5y=8. now multiply inside the brackets first. that is as far as i am taking you. you need to do a little of the work.
2006-06-22 07:52:21 · answer #3 · answered by dude 5 · 0⤊ 0⤋
solve first for x:
x = 3
then solve for y:
7(3) - 5y = 8
21 - 5y = 8
-5y = -13
y = 13/5
checking:
3-3 = 0
0 = 0 ------OK
7(3) - 5(13/5) = 8
21 - 13 = 8
8 = 8 ------OK
2006-06-22 07:57:59 · answer #4 · answered by Anonymous · 0⤊ 0⤋
when x=3 is the only posibility to satisfy the equation x-3=0
there fore we have to subtitute the value to the second equation
we get
7*3-5y=8
21-5y=8
5y=21-8
5y=13
y=13/5
2006-06-22 08:19:11 · answer #5 · answered by karthi 1 · 0⤊ 0⤋
the first one: x=3
3-3=0
the second one: x=4 and y=4
7x4-5x4=
=28-20=8
2006-06-22 07:53:07 · answer #6 · answered by Richmond 1 · 0⤊ 0⤋
x-3=0
x=3
21-5y=8
-5y=-21
y=4.2
2006-06-22 08:14:30 · answer #7 · answered by Raksha F 1 · 0⤊ 0⤋