How do you solve the following linear system: $3x-2y=7, 3x+4y=12$ ?

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Answer 1 · 2015-11-12T06:50:52

The solution for the system of equations is :
$color(blue)(y=5/6$

$color(blue)(x=(26)/(9)$

$color(blue)(3x)−2y=7$ ......equation $1$
$color(blue)(3x)+4y=12$ ......equation $2$

We can solve the system of equations by elimination

By subtracting equation $2$ from $1$ we can eliminate $color(blue)(3x$

$cancelcolor(blue)(3x)−2y=7$
$-cancelcolor(blue)(3x)-4y=-12$

$-6y=-5$

$y=(-5)/(-6)$

$color(blue)(y=5/6$

Now we substitute this value of $y$ in equation $1$ to find $x$ :

$3x−2y=7$

$3x=7+2y$

$3x=7+cancel2*color(blue)(5/cancel6$

$3x=7+5/3$

$3x=(21+5)/3$

$3x=(26)/3$

$x=(26)/(3*3)$

$color(blue)(x=(26)/(9)$

How do you solve the following linear system: 3x-2y=7, 3x+4y=123x-2y=7, 3x+4y=12?