Asked by td
5x-4y= -3
2x + 7y= 5
How to solve this linear system shown above by using elimination strategy ?
(5x-4y= -3) *2 ->10x - 12y= 6
(2x + 7y= 5) *5 -> 10 +35y= 25
___________________________________________
-47y = -19
y = -19/47
10x - 12y= 6
10x - 12(0.4) = 6
10x -4.8=6
x=1.08
I seem to get the final answer wrong. correct me if Im wrong overtime i verify. THanks
2x + 7y= 5
How to solve this linear system shown above by using elimination strategy ?
(5x-4y= -3) *2 ->10x - 12y= 6
(2x + 7y= 5) *5 -> 10 +35y= 25
___________________________________________
-47y = -19
y = -19/47
10x - 12y= 6
10x - 12(0.4) = 6
10x -4.8=6
x=1.08
I seem to get the final answer wrong. correct me if Im wrong overtime i verify. THanks
Answers
Answered by
Jai
You're doing the elimination method right. :) But there's just some errors in multiplying. In your first equation (5x - 4y = -3), you multiplied it by 2.
5x * 2 = 10x, yes that's right.
-4y * 2 = -8y, not -12y as you've typed.
-3 * 2 = -6, not 6 as you've typed.
5x * 2 = 10x, yes that's right.
-4y * 2 = -8y, not -12y as you've typed.
-3 * 2 = -6, not 6 as you've typed.
Answered by
td
oh sorry, I know I type it in wrong number. But even I tries in correct number. I still didn't get it right?
Answered by
Jai
Why is that? Anyway, let's solve for y. I'll multiply the first equation by -2 instead of 2:
(5x - 4y = -3) * -2 ---> -10x + 8y = 6
(2x + 7y = 5) * 5 ---> 10x + 35y = 25
Add them together:
-10x + 8y = 6
10x + 35y = 25
----------------------
43y = 31
y = 31/43
Now that you have a value for y, substitute this to either equation to get x. Let just choose the first equation:
5x - 4y = -3
5x - 4(31/43) = -3
x = ?
hope this helps~ `u`
(5x - 4y = -3) * -2 ---> -10x + 8y = 6
(2x + 7y = 5) * 5 ---> 10x + 35y = 25
Add them together:
-10x + 8y = 6
10x + 35y = 25
----------------------
43y = 31
y = 31/43
Now that you have a value for y, substitute this to either equation to get x. Let just choose the first equation:
5x - 4y = -3
5x - 4(31/43) = -3
x = ?
hope this helps~ `u`
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