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lemjay eNotes educator| Certified Educator

`13x - 5y = 10`      (Let this be EQ1.)

`3x - 2y =14`       (Let this be EQ2.)

To solve for values of x and y, elimination method can be applied. So, let's try to eliminate the variable y.

To eliminate y, multiply EQ1 by 2 and EQ3 by -5. So, that the coefficients of y become the same but with different signs.

2*EQ1: `26x - 10y=20`

-5*EQ2: `-15x + 10y = -70`

Then, add the products of 2EQ1 and -5EQ2.

          `26x - 10y =20`
`(+)` `-15x+10y=-70`
         `11x`              `=`  `-50`

And, isolate x.



Then, substitute this value of x to either of the original equation. So, plug-in x=-50/11 to EQ2.










Hence, the solution to the given system of equations is `(-50/11,-152/11)` .

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