How do you solve the following linear system $3 x - 7 y = 1, - 3 x + 5 y = 1 ?$ $3x-7y=1, -3x+5y=1?$ ?

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How do you solve the following linear system $3 x - 7 y = 1, - 3 x + 5 y = 1 ?$ $3x-7y=1, -3x+5y=1?$ ?

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Answer 1 · 2017-03-07T08:53:07

$x = - 2$ $x=-2$ and $y = - 1$ $y=-1$

Explanation:

$3 x - 7 y = 1$ $3x-7y=1$
$- 3 x + 5 y = 1$ $-3x+5y=1$

From the first equation, determine a temporary value for $3 x$ $3x$ .

$3 x - 7 y = 1$ $3x-7y=1$

Add $7 y$ $7y$ to each side.

$3 x = 7 y + 1$ $3x=7y+1$

In the second equation, substitute $3 x$ $3x$ with $(7 y + 1)$ $color(red)((7y+1))$ .

$- 3 x + 5 y = 1$ $-3x+5y=1$

$- (7 y + 1) + 5 y = 1$ $-color(red)((7y+1))+5y=1$

Open the brackets and simplify. The product of a negative and a positive is a negative.

$- 7 y - 1 + 5 y = 1$ $-7y-1+5y=1$

$- 2 y - 1 = 1$ $-2y-1=1$

Add $1$ $1$ to both sides.

$- 2 y = 2$ $-2y=2$

Divide both sides by $- 2$ $-2$ .

$y = - 1$ $y=-1$

In the first equation, substitute $y$ $y$ with $- 1$ $color(blue)(-1)$ .

$3 x - 7 y = 1$ $3x-7y=1$

$3 x - 7 (- 1) = 1$ $3x-7color(blue)((-1))=1$

Open the brackets and simplify. The product of two negatives is a positive.

$3 x + 7 = 1$ $3x+7=1$

Subtract $7$ $7$ from each side.

$3 x = - 6$ $3x=-6$

Divide both sides by $3$ $3$ .

$x = - 2$ $x=-2$

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How do you solve the following linear system $3 x - 7 y = 1, - 3 x + 5 y = 1 ?$ $3x-7y=1, -3x+5y=1?$ ?

1 Answer

Explanation:

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How do you solve the following linear system 3x-7y=1, -3x+5y=1? 3x−7y=1,−3x+5y=1?3x-7y=1, -3x+5y=1? ?

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Explanation:

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How do you solve the following linear system $3 x - 7 y = 1, - 3 x + 5 y = 1 ?$ $3x-7y=1, -3x+5y=1?$ ?