How do you solve the system $x/2-y/3=5/6$ and $x/5-y/4=71/10$ ?

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Answer 1 · 2016-03-21T07:22:42

Scale and combine the equations to eliminate $y$ and solve for $x$ , then substitute that value of $x$ back in to solve for $y$ and find:

${ (x = -37), (y = -58) :}$

Given:

${ (x/2-y/3=5/6), (x/5-y/4=71/10) :}$

Multiply the first equation by $6$ and the second by $20$ to get:

${ (3x-2y=5), (4x-5y=142) :}$

To eliminate $y$ , multiply the first of these by $5$ , the second by $2$ and subtract the second from the first...

${(15x-10y=25),(8x-10y=284):}$

$=> 7x = 25-284 = -259$

Then divide both ends by $7$ to find:

$x = -259/7 = -37$

Substitute this value of $x$ into the equation $3x-2y=5$ to get:

$-111-2y=5$

Add $111$ to both sides to get:

$-2y=116$

Divide both sides by $-2$ to get:

$y = -58$

How do you solve the system x/2-y/3=5/6x/2-y/3=5/6 and x/5-y/4=71/10x/5-y/4=71/10?