Question

Solve the system by the elimination method. 3x - 2y - 7 = 0 5x + y - 3 = 0 To eliminate y, the LCM is 2. Which of the following is the resulting equations? 3x - 2y - 7 = 0 5x + y - 3 = 0 3x - 2y - 7 = 0 -10x - 2y + 6 = 0 3x - 2y - 7 = 0 10x + 2y - 6 = 0

Answer 1

Answer:

$x=1$ and $y=-2$.

Step-by-step explanation:

We have been given a system of equations. We are asked to simplify our given system by the elimination method.

$3x-2y-7=0...(1)$

$5x+y-3=0...(2)$

First of all, we will multiply equation (2) by 2 as shown below:

$2*5x+2*y-2*3=0...(2)$

$10x+2y-6=0...(2)$

Upon adding equation (1) and (2), we will get:

$(3x-2y-7=0)+(10x+2y-6=0)\rightarrow 3x+10x-2y+2y-7-6=0$

$13x-13=0$

$13x-13+13=0+13$

$13x=13$

$\frac{13x}{13}=\frac{13}{13}$

$x=1$

Upon substituting $x=1$ in equation (2), we will get:

$5*1+y-3=0$

$5+y-3=0$

$2+y=0$

$2-2+y=0-2$

$y=-2$

Therefore, the solution for our given system of equation is $x=1$ and $y=-2$.

Answer 2

Next time, please share your system of linear equations by typing only one equation per line:

3x - 2y - 7 = 0 5x + y - 3 = 0  NO

3x - 2y - 7 = 0
5x + y - 3 = 0                          YES

Mult. the 2nd equation by 2 so as to obtain 2y, which will be cancelled out by  - 2y in the first equation:

  3x - 2y - 7 = 0
2(5x + y - 3 = 0)

Then:

  3x - 2y - 7 = 0
10x +2y - 6 = 0
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13x - 13 = 0, so that x = 1.  Find y by subbing 1 for x in either of the 2 given equations.