Explanation:
You eliminate a variable by combining the equations in such a way that one of the variables ends up with a coefficient of zero. In general, you multiply each equation by a number chosen so that when the two products are added, one of the coefficients becomes zero.
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Here, we notice that the coefficient of y in the first equation is -2 times the coefficient of y in the second equation. So, if we multiply the second equation by 2 and add that result to the first equation, the y terms will cancel:
2(5x +y) +(3x -2y) = 2(4) +(10)
10x +2y +3x -2y = 8 +10 . . . . eliminate parentheses
13x = 18 . . . . . . . . . . . . . . . . . . collect terms. (The y-term is eliminated.)
x = 18/13 . . . . divide by the coefficient of x
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We can also eliminate the x-variable, but there is a little more work involved. We can multiply the first equation by -5 and add that to the second equation after it is multiplied by 3. You will notice these multipliers (-5, 3) are each the coefficient of the variable in the other equation, with one of them negated.
Following this process, we have ...
-5(3x -2y) +3(5x +y) = -5(10) +3(4)
-15x +10y +15x +3y = -50 +12 . . . . . eliminate parentheses
13y = -38 . . . . . . . . . . . . . . . . . . . . . . collect terms*
y = -38/13 . . . . divide by the coefficient of y
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* You will notice the coefficient of y is positive. This is the result of choosing multipliers of -5 and 3, rather than 5 and -3. A little planning ahead can help reduce errors.
