How do solve the following linear system?: $4 x + 3 y = 29, 2 x - 5 y = 3$ $4x+3y=29, 2x - 5y=3$ ?

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How do solve the following linear system?: $4 x + 3 y = 29, 2 x - 5 y = 3$ $4x+3y=29, 2x - 5y=3$ ?

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Answer 1 · 2018-06-17T19:07:18

$x = \frac{77}{13} = 5 \frac{12}{13}$ $x=77/13 = 5 12/13$
$y = \frac{23}{13} = 1 \frac{10}{13}$ $y=23/13=1 10/13$

Explanation:

We have
1) $4 x + 3 y = 29$ $4x+3y=29$
2) $2 x - 5 y = 3$ $2x-5y=3$
Double each term in 2):
3) $4 x - 10 y = 6$ $4x-10y=6$

subtract 3) from 1):
$4 x - 4 x + 3 y + 10 y = 29 - 6 = 23$ $4x-4x +3y+10y =29-6=23$
4) $13 y = 23$ $13y=23$
4) gives that
5) $y = \frac{23}{13} = 1 \frac{10}{13}$ $y=23/13=1 10/13$

Insert this in 2):
6) $2 x - \frac{115}{13} = 3$ $2x-115/13=3$
$2 x = \frac{115 + 39}{13} = \frac{154}{13}$ $2x=(115+39)/13=154/13$
$x = \frac{77}{13} = 5 \frac{12}{13}$ $x=77/13 = 5 12/13$

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How do solve the following linear system?: $4 x + 3 y = 29, 2 x - 5 y = 3$ $4x+3y=29, 2x - 5y=3$ ?

1 Answer

Explanation:

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How do solve the following linear system?: 4x+3y=29, 2x - 5y=3 4x+3y=29,2x−5y=3 4x+3y=29, 2x - 5y=3 ?

1 Answer

Explanation:

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How do solve the following linear system?: $4 x + 3 y = 29, 2 x - 5 y = 3$ $4x+3y=29, 2x - 5y=3$ ?