Solve these system of linear equations:
5x-y=10}
4x-2y=20}

AND:

3x+4y=12}
2x-y=-10}

WITH STEPS

$$5x-y=10 \\ 4x-2y=20 \ \ \ |\hbox{divide both sides by (-2)} \\ \\ 5x-y=10 \\ \underline{-2x+y=-10} \ \ \ |\hbox{add by sides} \\ 5x-2x-y+y=10-10 \\ 3x=0 \\ x=0 \\ \\ \hbox{substitute 0 for x in one of the equations:} \\ 5 \times 0 -y=10 \\ -y=10 \\ y=-10 \\ \\ \hbox{the answer:} \\ x=0 \\ y=-10$$

$$3x+4y=12 \\ 2x-y=-10 \ \ \ |\hbox{multiply both sides by 4} \\ \\ 3x+4y=12 \\ \underline{8x-4y=-40} \ \ \ |\hbox{add by sides} \\ 3x+8x+4y-4y=12-40 \\ 11x=-28 \\ x=-\frac{28}{11} \\ x=-2 \frac{6}{11} \\ \\ \hbox{substitute -28/11 for x in one of the equations:} \\ 2 \times (-\frac{28}{11})-y=-10 \\ -\frac{56}{11}-y=-\frac{110}{11} \\ -y=-\frac{110}{11}+\frac{56}{11} \\ -y=-\frac{54}{11} \\ y=\frac{54}{11} \\ y=4 \frac{10}{11} \\ \\ \hbox{the answer:} \\ x=-2\frac{6}{11} \\ y=4 \frac{10}{11}$$

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