How do you solve the system of equations: y = 3x - 4 and 2x - y = 1?

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How do you solve the system of equations: y = 3x - 4 and 2x - y = 1?

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Answer 1 · 2015-09-17T16:21:00

I found:
$x = 3$ $x=3$
$y = 5$ $y=5$

Explanation:

You can substitute the value of $y$ $y$ from the first equation into the second and solve for $x$ $x$ as:
$2 x - [3 x - 4] = 1$ $2x-color(red)([3x-4])=1$
$2 x - 3 x + 4 = 1$ $2x-3x+4=1$
$- x = - 3$ $-x=-3$
$x = 3$ $x=3$
substitute back this value into the first equation to find $y$ $y$ :
$y = 3 \cdot (3) - 4 = 5$ $y=3*(3)-4=5$

Answer 2 · 2015-09-17T16:33:16

Refer to explanation

Explanation:

We have two equations

$y = 3 x - 4$ $y=3x-4$ and $y = 2 x - 1$ $y=2x-1$ Multiply the first with 2 and the second with three and substract them

$2 y - 3 y = 2 \cdot (3 x - 4) - 3 (2 x - 1) \Rightarrow - y = - 5 \Rightarrow y = 5$ $2y-3y=2*(3x-4)-3(2x-1)=>-y=-5=>y=5$ and $x = 3$ $x=3$

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How do you solve the system of equations: y = 3x - 4 and 2x - y = 1?

2 Answers

Explanation:

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