How do solve the following linear system?: $x - 3 y = - 6, 10 x - 7 y = - 18$ $x - 3y = -6 , 10x - 7y = -18$ ?

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Answer 1 · 2016-01-27T14:13:19

${x, y} = {- \frac{12}{23}, \frac{42}{23}}$ ${x,y}={-12/23,42/23}$

Solve by elimination

$x - 3 y = - 6$ $x-3y=-6$
$10 x - 7 y = - 18$ $10x-7y=-18$

We can eliminate $10 x$ $10x$ from the second equation by $x$ $x$ in the first equation if we multiply $x$ $x$ by $- 10$ $-10$ to get $- 10 x$ $-10x$

$\to - 10 (x - 3 y = - 6)$ $rarr-10(x-3y=-6)$

$\to - 10 x + 30 y = 60$ $rarr-10x+30y=60$

Now add both of the equations

$\to (- 10 x + 30 y = 60) + (10 x - 7 y = - 18)$ $rarr(-10x+30y=60)+(10x-7y=-18)$

$\to 23 y = 42$ $rarr23y=42$

$y = \frac{42}{23}$ $y=42/23$

Now substitute the value of y to the first equation

$x - 3 (\frac{42}{23}) = - 6$ $x-3(42/23)=-6$

$x - \frac{126}{23} = - 6$ $x-126/23=-6$

$x = - 6 + \frac{126}{23} = \frac{- 138 + 126}{23} = - \frac{12}{23}$ $x=-6+126/23=(-138+126)/23=-12/23$

So, ${x, y} = {- \frac{12}{23}, \frac{42}{23}}$ ${x,y}={-12/23,42/23}$

How do solve the following linear system?: x−3y=−6,10x−7y=−18 x - 3y = -6 , 10x - 7y = -18 ?