How do you solve the following linear system: $4 x + 3 y = 8, x - 2 y = 13$ $4x+3y=8 , x-2y=13$ ?

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How do you solve the following linear system: $4 x + 3 y = 8, x - 2 y = 13$ $4x+3y=8 , x-2y=13$ ?

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Answer 1 · 2016-01-28T11:38:59

$(x, y) = (5, - 4)$ $(x,y)=(5,-4)$

Explanation:

$4 x + 3 y = 8$ $4x+3y=8$

$x - 2 y = 13$ $x-2y=13$

In the second equation we can find the value of $x$ $x$ by transposing $2 y$ $2y$ to the other side of the equation and we can substitute the value of $x$ $x$ to the other equation

Solve for second equation:

$\to x - 2 y = 13$ $rarrx-2y=13$

Add $2 y$ $2y$ both sides:

$\to x = 13 + 2 y$ $rarrx=13+2y$

Substitute the value of $x$ $x$ to the first equation:

$\to 4 (13 + 2 y) + 3 y = 8$ $rarr4(13+2y)+3y=8$

$\to (52 + 8 y) + 3 y = 8$ $rarr(52+8y)+3y=8$

$\to 52 + 8 y + 3 y = 8$ $rarr52+8y+3y=8$

$\to 52 + 11 y = 8$ $rarr52+11y=8$

$\to 11 y = 8 - 52$ $rarr11y=8-52$

$\to 11 y = - 44$ $rarr11y=-44$

$\to y = - \frac{44}{11} = - 4$ $rarry=-44/11=-4$

So,substitute value of y to second equation:

$\to x - 2 (- 4) = 13$ $rarrx-2(-4)=13$

$\to x - (- 8) = 13$ $rarrx-(-8)=13$

$\to x + 8 = 13$ $rarrx+8=13$

$\to x = 13 - 8 = 5$ $rarrx=13-8=5$

So, $(x, y) = (5, - 4)$ $(x,y)=(5,-4)$

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How do you solve the following linear system: $4 x + 3 y = 8, x - 2 y = 13$ $4x+3y=8 , x-2y=13$ ?

1 Answer

Explanation:

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How do you solve the following linear system: $4 x + 3 y = 8, x - 2 y = 13$ $4x+3y=8 , x-2y=13$ ?