The Elimination Method

The idea in a nut­shell#

The elim­i­na­tion method gets rid of one vari­able by adding or sub­tract­ing the two equa­tions, after first scal­ing them so that the vari­able has match­ing coef­fi­cients. When no vari­able has coef­fi­cient $1$, it is usu­ally quicker than sub­sti­tu­tion, and it keeps frac­tions away until the very last step.

The method#

1. Mul­ti­ply one or both equa­tions by con­stants so that one vari­able has the same coef­fi­cient in both (or the same size with oppo­site signs).
2. Sub­tract (or add) the equa­tions to elim­i­nate that vari­able.
3. Solve the result­ing equa­tion in one vari­able.
4. Sub­sti­tute back into either orig­i­nal equa­tion to find the other vari­able.

As always, ver­ify in both orig­i­nal equa­tions.

Worked exam­ple 1, add#

Solve $3x + 4y = 10$ and $2x - 2y = 2$.

Elim­i­nate $y$. Mul­ti­ply the sec­ond equa­tion by $2$ so the $y$-coef­fi­cients are $4$ and $-4$:

$$3x + 4y = 10 \qquad (1)$$
$$4x - 4y = 4 \qquad (2)$$

Add (1) and (2): $7x = 14$, so $x = 2$.

Sub­sti­tute into $2x - 2y = 2$: $4 - 2y = 2$, so $y = 1$.

Ver­ify: $6 + 4 = 10$ ✓ and $4 - 2 = 2$ ✓.

Worked exam­ple 2, scale both#

Solve $5x + 3y = 19$ and $3x - 2y = 3$.

This time nei­ther vari­able lines up. To elim­i­nate $y$, make its coef­fi­cients $6$ and $-6$ by mul­ti­ply­ing the first equa­tion by $2$ and the sec­ond by $3$.

$$10x + 6y = 38$$
$$9x - 6y = 9$$

Add: $19x = 47$, so $\displaystyle x = \frac{47}{19}$. A frac­tion! That is a sig­nal to re-check your arith­metic, not a rea­son to assume you went wrong. Here $10x + 9x = 19x$ and $38 + 9 = 47$ are both right.

To be dou­bly sure, elim­i­nate $x$ instead: mul­ti­ply the first by $3$ and the sec­ond by $5$.

$$15x + 9y = 57$$
$$15x - 10y = 15$$

Sub­tract: $19y = 42$, so $\displaystyle y = \frac{42}{19}$. Also a frac­tion. The solu­tion gen­uinely is frac­tional: $\displaystyle x = \frac{47}{19}$, $\displaystyle y = \frac{42}{19}$. Ver­ify: $\displaystyle 5 \cdot \frac{47}{19} + 3 \cdot \frac{42}{19} = \frac{235 + 126}{19} = \frac{361}{19} = 19$ ✓.

The moral: an ugly answer is not proof of an error. Re-check the arith­metic, and then trust your work.

Worked exam­ple 3, signs#

Solve $4x - 3y = 11$ and $x + 3y = 4$.

The $y$-coef­fi­cients are already $-3$ and $+3$. Add the equa­tions: $5x = 15$, $x = 3$. Then $3 + 3y = 4$, $\displaystyle y = \frac{1}{3}$.

Ver­ify: $12 - 1 = 11$ ✓ and $3 + 1 = 4$ ✓.

Whether you add or sub­tract depends only on the signs: same sign, sub­tract; oppo­site signs, add. Get this back­wards and you dou­ble the vari­able instead of remov­ing it, but you'll soon notice, because noth­ing can­cels.

When both vari­ables van­ish#

Just like sub­sti­tu­tion, elim­i­na­tion warns you about a non-unique case by wip­ing out both vari­ables at once.

$2x + 3y = 6$ and $4x + 6y = 12$: dou­ble the first and sub­tract from the sec­ond: $0 = 0$. True, infi­nitely many solu­tions.

$2x + 3y = 6$ and $4x + 6y = 15$: same step gives $0 = 3$. False, no solu­tion.

Prac­tice#

Solve by elim­i­na­tion and ver­ify.

1. $x + y = 9$, $x - y = 1$.
2. $2x + 3y = 13$, $2x - y = 1$.
3. $3x + 2y = 12$, $2x + 5y = 19$.
4. $5x - 2y = 4$, $3x + 4y = 18$.
5. $7x + 2y = 20$, $3x - 4y = -6$.
6. $6x - 4y = 10$, $3x - 2y = 7$.

Answers#