How do you solve $5 x + 8 y = 28$ $5x+8y=28$ and $- 8 + y = 38$ $-8+y=38$ using substitution?

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How do you solve $5 x + 8 y = 28$ $5x+8y=28$ and $- 8 + y = 38$ $-8+y=38$ using substitution?

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Answer 1 · 2018-07-12T12:44:54

$(x, y) \to (- 68, 46)$ $(x,y)to(-68,46)$

Explanation:

$5 x + 8 y = 28 \to (1)$ $5x+8y=28to(1)$

$- 8 + y = 38 \to (2)$ $-8+y=38to(2)$

$simplify equation (2) by adding 8 to both sides$ $"simplify equation "(2)" by adding 8 to both sides"$

$y = 38 + 8 = 46$ $y=38+8=46$

$substitute y = 46 into equation (1)$ $"substitute "y=46" into equation "(1)$

$5 x + (8 \times 46) = 28$ $5x+(8xx46)=28$

$5 x + 368 = 28$ $5x+368=28$

$subtract 368 from both sides$ $"subtract 368 from both sides"$

$5 x = 28 - 368 = - 340$ $5x=28-368=-340$

$divide both sides by 5$ $"divide both sides by 5"$

$x = \frac{- 340}{5} = - 68$ $x=(-340)/5=-68$

$solution is (x, y) \to (- 68, 46)$ $"solution is "(x,y)to(-68,46)$

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How do you solve $5 x + 8 y = 28$ $5x+8y=28$ and $- 8 + y = 38$ $-8+y=38$ using substitution?

1 Answer

Explanation:

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How do you solve $5 x + 8 y = 28$ $5x+8y=28$ and $- 8 + y = 38$ $-8+y=38$ using substitution?