What is the solution to the system of equations using the linear combination method? {3x+y=42x+y=7 (−1, 7) (−3, 12) (−3, 13) (0, 4)

(-3, 13)

We can subtract the second equation from the first to eliminate the y-variable.

... (3x +y) -(2x +y) = (4) -(7)

... x = -3 . . . . . simplify

Using either equation, we can find y.

... 3x +y = 4 . . . . the first equation

... y = 4 -3x = 4 -3(-3) = 13 . . . . . the first equation rearranged by subtracting 3x

The solution is (x, y) = (-3, 13).

Comment on the solution

The coefficient of y is 1 in both equations. This means that multiplying one of the equations by -1 and adding that result to the other will make the y-terms add to zero (cancel). This is the goal of "linear combination": cancel one of the variables.

Another combination we could have used is multiplying the first equation by -2/3 and adding to the second. This would cancel the x-terms.

Generally, we prefer to work with integers, so if we actually wanted to cancel x-terms, we would likely multiply the second equation by 3 and the first equation by -2, then add the results of those operations. This would give

... 3(2x +y) -2(3x +y) = 3(7) -2(4)

... y = 13 . . . . . simplify

You will note that the choice of multipliers made 3(2x) the opposite of -2(3x), so when these products are added, the result is zero. That is the reason behind the choice of multipliers. As a starting point, you can choose the multipliers to be the cofficient of the variable in one equation, and the opposite of the coefficient of the variable in the other equation.

Here, x-coefficients are 3 and 2; the multipliers we chose were -2 and 3. (The reason for not using 2 and -3 is that we want a positive coefficient for y when we're done simplifying.)

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$12. that is my answer.