How do you solve $y = 4 x - 3, 2 y - 3 x = 4$ $y=4x-3, 2y-3x=4$ ?

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How do you solve $y = 4 x - 3, 2 y - 3 x = 4$ $y=4x-3, 2y-3x=4$ ?

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Answer 1 · 2017-01-11T20:54:52

$(2, 5)$ $(2,5)$

Explanation:

Since we are given $y = 4 x - 3$ $y=4x-3$ we can substitute this into the other equation and solve for x.

$\Rightarrow 2 (4 x - 3) - 3 x = 4$ $rArr2(color(red)(4x-3))-3x=4$

distributing.

$8 x - 6 - 3 x = 4$ $8x-6-3x=4$

add 6 to both sides and simplify.

$5xcancel(-6)cancel(+6)=4+6$

$rArr5x=10$

To solve for x, divide both sides by 5

$(cancel(5) x)/cancel(5)=10/5rArrx=2$

We can now substitute x = 2 into y = 4x - 3 to find y.

$x=2toy=(4xx2)-3=8-3=5$

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How do you solve $y = 4 x - 3, 2 y - 3 x = 4$ $y=4x-3, 2y-3x=4$ ?

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Explanation:

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How do you solve y=4x-3, 2y-3x=4y=4x−3,2y−3x=4y=4x-3, 2y-3x=4?

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Explanation:

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How do you solve $y = 4 x - 3, 2 y - 3 x = 4$ $y=4x-3, 2y-3x=4$ ?