Answer:
- multipliers: 3, 2 for subtraction; or 3, -2 for addition.
- solution: (x, y) = (-4 1/6, 7 2/3)
Explanation:
You want to solve the given system of equations by eliminating the y-term. You also want to know what constants are used in the process.
Elimination
The y-term can be eliminated from these equations by combining them in such a way as to may the coefficient of y be zero. This can be done an infinite number of ways.
One way that works reasonably well is to identify the coefficients of y, and multiply each equation by the y-coefficient in the other equation. The resulting equations can be subtracted one from the other to eliminate the y-terms, which now have identical coefficients. You can choose the equation with the smaller x-coefficient to subtract from the other one.
Application
The y-term coefficients are 2 and 3.
Multiply the first equation by 3, and the second by 2.
3 × (2x +2y) = 3 × 7 ⇒ 6x +6y = 21
2 × (6x +3y) = 2 × (-2) ⇒ 12x +6y = -4
The x-coefficients are 6 and 12. We choose to subtract the first equation (with x-coefficient 6) from the second equation:
(12x +6y) -(6x +6y) = (-4) -(21)
6x = -25 . . . . . . . . simplified
x = -25/6 = -4 1/6 . . . . . divide by 6
Substituting for x in the first equation, we have ...
2(-25/6) +2y = 7
2y = 7 +25/3 . . . . . add 25/3
2y = 46/3 . . . . . . . .combine terms
y = 23/3 = 7 2/3 . . . . divide by 2
The solution to the equations is (x, y) = (-4 1/6, 7 2/3).
Check
Since these numbers are not integers, we choose to check them in the original equations. We used the first equation to find y, so we'll use the second equation for checking.
6(-4 1/6) +3(7 2/3) = -25 +23 = -2 . . . . . . as required
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Additional comment
If you solve these by addition, rather than subtraction, you want to choose multipliers that result in opposite coefficients, so they add to zero. Essentially, you use the opposite coefficients, as described above, but you negate one of them.
Above, we chose the subtraction in a way that resulted in the x-coefficient being positive. If you arbitrarily choose the signs of the multipliers (always negating the second one, for example), the combined equations may make the x-coefficient negative. That works, too, but may require more mental effort to deal with the minus sign.
Of course, you can use the x-coefficients and eliminate the x-terms.
Or, you can use a single multiplier that corresponds to the ratio of the coefficients. For example the ratio of the 2nd y-coefficient to the first one is 3/2. Multiplying the first equation by this will give the y-term a coefficient of 3, matching the y-coefficient in the 2nd equation. Then the subtraction can proceed as already described.
(6x +3y) -3/2(2x +2y) = (-2) -3/2(7)
3x = -25/2 . . . . . simplified; y is gone.