How do you solve the following system: $11y + 3x = 4, -6y - 2x = 0 -3$ ?

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Answer 1 · 2016-07-25T18:27:35

The Soln. $x=9/4, y=-1/4$ .

$11y+3x=4................(1)$

$-6y-2x=-3...........(2)$ .

Multiplying $(1)$ by $2$ , we get, $22y+6x=8... ... (1star)$

Multiplying $(2)$ by $3$ , we get, $-18y-6x=-9... ...(2star)$

Adding $(1star) and (2star)$ , we get,

$22ycancel(+6x)-18ycancel(-6x)=8-9$

$:. 4y=-1 rArr y=-1/4$ .

Sub.ing $y=-1/4$ in $(2)$ , we get,

$-6(-1/4)-2x=-3, i.e., -2x=-3-3/2=-9/2$

$:. x=9/4$ .

Hence, the Soln. $x=9/4, y=-1/4$ .

How do you solve the following system: 11y + 3x = 4, -6y - 2x = 0 -3 11y + 3x = 4, -6y - 2x = 0 -3 ?