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secondary 2 | Maths
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How to do this question using elimination method. Thanks.
Date Posted: 3 years ago
Views: 255
5x - 4y = 23 —> Eq 1
2x - 7y = 11 —> Eq 2
We make the “x” have the same number.
Eq 1 multiplied by 2 gives
10x - 8y = 46 —> Eq 3
Eq 2 multiplied by 5 gives
10x - 35 = 55 —> Eq 4
Eq 3 - Eq 4 gives
(10x - 8y) - (10x - 35y) = 46 - 55
27y = -9
y = -1/3
Sub y = -1/3 into Eq 1,
5x - 4 (-1/3) = 23
5x + 4/3 = 23
5x = 65/3
x = 13/3
2x - 7y = 11 —> Eq 2
We make the “x” have the same number.
Eq 1 multiplied by 2 gives
10x - 8y = 46 —> Eq 3
Eq 2 multiplied by 5 gives
10x - 35 = 55 —> Eq 4
Eq 3 - Eq 4 gives
(10x - 8y) - (10x - 35y) = 46 - 55
27y = -9
y = -1/3
Sub y = -1/3 into Eq 1,
5x - 4 (-1/3) = 23
5x + 4/3 = 23
5x = 65/3
x = 13/3
See 1 Answer
done
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clear
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Hope this helps
Date Posted:
3 years ago
he's asking for elimination, this is substitution. Presentation wise can be shown better
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