How do you solve the system $\frac{1}{3} x - \frac{1}{12} y = - 7$ $1/3 x - 1/12 y = -7$ and $\frac{1}{3} x + \frac{1}{6} y = 11$ $1/3 x + 1/6 y = 11$ ?

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How do you solve the system $\frac{1}{3} x - \frac{1}{12} y = - 7$ $1/3 x - 1/12 y = -7$ and $\frac{1}{3} x + \frac{1}{6} y = 11$ $1/3 x + 1/6 y = 11$ ?

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Answer 1 · 2018-03-16T05:54:47

x = -3 and y = 72

Explanation:

So we are given with $2$ $2$ equations

equation 1 is $\frac{1}{3} x - \frac{1}{12} y$ $1/3x - 1/12y$ = -7

equation 2 is $\frac{1}{3} x + \frac{1}{6} y$ $1/3x + 1/6y$ = 11

equation 1 can be written as $4 x - y = - 84$ $4x - y = -84$
(multiplied the whole equation by 12)

equation 2 can be written as $2 x + y = 66$ $2x + y = 66$
(multiplied whole equation by 6)

now we add both the equations

==> $4 x - y + 2 x + y = - 84 + 66$ $4x - y + 2x + y = -84 + 66$
==> $6 x = - 18$ $6x = -18$

==> $x = - \frac{18}{6}$ $x = -18/6$

==> $x = - 3$ $x = -3$

now we substitute $x = - 3$ $x = -3$ in any one of the equations
we will get $y = 72$ $y = 72$

Answer 2 · 2018-03-19T13:56:02

$x = - 3 and y = 72$ $x = -3 and y =72$

Explanation:

Both equations have the same $x$ $x$ - coeffiecients.

Rewrite both equations:

$\frac{1}{3} x = \frac{1}{12} y - 7 and \frac{1}{3} x = - \frac{1}{6} y + 11$ $1/3x = 1/12y -7" "and" "1/3x= -1/6y +11$

$\frac{1}{3} x = \frac{1}{3} x$ $1/3x = 1/3x$ , so equate the right sides of the equations.

$\frac{1}{12} y - 7 = - \frac{1}{6} y + 11 \leftarrow$ $1/12y -7 = -1/6y +11" "larr$ multiply by $12$ $12$

$y - 84 = - 2 y + 132$ $y-84 = -2y+132$

$y + 2 y = 132 + 84$ $y +2y = 132+84$

$3 y = 216$ $3y = 216$

$y = 72$ $y = 72$

Substitute to find $x$ $x$

$\frac{1}{3} x = \frac{1}{12} (72) - 7$ $1/3x = 1/12 (72)-7$

$\frac{1}{3} x = 6 - 7$ $1/3x = 6-7$

$\frac{1}{3} x = - 1$ $1/3x = -1$

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

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How do you solve the system $\frac{1}{3} x - \frac{1}{12} y = - 7$ $1/3 x - 1/12 y = -7$ and $\frac{1}{3} x + \frac{1}{6} y = 11$ $1/3 x + 1/6 y = 11$ ?

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How do you solve the system 1/3 x - 1/12 y = -713x−112y=−71/3 x - 1/12 y = -7 and 1/3 x + 1/6 y = 1113x+16y=111/3 x + 1/6 y = 11?

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How do you solve the system $\frac{1}{3} x - \frac{1}{12} y = - 7$ $1/3 x - 1/12 y = -7$ and $\frac{1}{3} x + \frac{1}{6} y = 11$ $1/3 x + 1/6 y = 11$ ?