The elimination method

The sec­ond method, elim­i­na­tion, is usu­ally the quick­est by hand. Instead of mak­ing one vari­able the sub­ject, you mul­ti­ply the equa­tions so that one vari­able has the same coef­fi­cient in both, then add or sub­tract the equa­tions so that vari­able dis­ap­pears, or is elim­i­nated. What remains is an equa­tion in one vari­able.

The method#

1. Write both equa­tions with the $x$ terms, the $y$ terms and the num­bers lined up in columns. Num­ber them (1) and (2).
2. Choose a vari­able to elim­i­nate. Mul­ti­ply one or both equa­tions by num­bers that make its coef­fi­cients equal in size.
3. If those coef­fi­cients have oppo­site signs, add the equa­tions. If they have the same sign, sub­tract.
4. Solve the one-vari­able equa­tion that remains.
5. Sub­sti­tute that value in either orig­i­nal equa­tion to find the other vari­able, then check.

Sub­tract­ing by chang­ing signs#

Sub­tract­ing a whole equa­tion is where slips hap­pen. A safe way is to change the sign of every term in the lower equa­tion and then add. Write the new signs above the old ones before you add. Sub­tract­ing $6x - 9y = 21$ is the same as adding $-6x + 9y = -21$.

Exam­ple 1#

Solve $3x + 2y = 5$ and $2x - 3y = 7$.

$$\begin{aligned} 3x + 2y &= 5 \quad \text{(1)} \\ 2x - 3y &= 7 \quad \text{(2)} \end{aligned}$$

To elim­i­nate $x$, make both $x$ coef­fi­cients equal to $6$: mul­ti­ply (1) by $2$ and (2) by $3$.

$$\begin{aligned} \text{(1)} \times 2: \quad 6x + 4y &= 10 \\ \text{(2)} \times 3: \quad 6x - 9y &= 21 \end{aligned}$$

Both $x$ terms are $+6x$, the same sign, so sub­tract. Change the signs of the lower line, which becomes $-6x + 9y = -21$, and add:

$$\displaystyle \begin{aligned} (6x - 6x) + (4y + 9y) &= 10 - 21 \\ 13y &= -11 \\ y &= -\frac{11}{13} \end{aligned}$$

Sub­sti­tute in (1):

$$\displaystyle \begin{aligned} 3x + 2\left(-\frac{11}{13}\right) &= 5 \\ 3x - \frac{22}{13} &= 5 \\ 3x &= 5 + \frac{22}{13} = \frac{65 + 22}{13} \\ 3x &= \frac{87}{13} \\ x &= \frac{29}{13} \end{aligned}$$

$$\displaystyle \therefore (x, y) = \left(\frac{29}{13}, -\frac{11}{13}\right)$$

This is exactly the answer the sub­sti­tu­tion method gave in the last les­son, as it must be.

The same pair, elim­i­nat­ing y instead#

You may elim­i­nate either vari­able. To remove $y$, make both $y$ coef­fi­cients $6$: mul­ti­ply (1) by $3$ and (2) by $2$.

$$\begin{aligned} \text{(1)} \times 3: \quad 9x + 6y &= 15 \\ \text{(2)} \times 2: \quad 4x - 6y &= 14 \end{aligned}$$

Now the $y$ terms have oppo­site signs, $+6y$ and $-6y$, so sim­ply add:

$$\displaystyle \begin{aligned} 13x &= 29 \\ x &= \frac{29}{13} \end{aligned}$$

Same answer again. Choose whichever vari­able needs the smaller mul­ti­pli­ers, or the one whose signs are already oppo­site, so that you can add.

Exam­ple 2#

Solve $7x - 5y = 2$ and $x + 2y = 3$.

$$\begin{aligned} 7x - 5y &= 2 \quad \text{(1)} \\ x + 2y &= 3 \quad \text{(2)} \end{aligned}$$

Make the $x$ coef­fi­cients equal to $7$: leave (1) as it is and mul­ti­ply (2) by $7$.

$$\begin{aligned} \text{(1)} \times 1: \quad 7x - 5y &= 2 \\ \text{(2)} \times 7: \quad 7x + 14y &= 21 \end{aligned}$$

Same sign on $7x$, so sub­tract: change the lower line to $-7x - 14y = -21$ and add.

$$\begin{aligned} -19y &= -19 \\ y &= 1 \end{aligned}$$

Sub­sti­tute in (2):

$$\begin{aligned} x + 2(1) &= 3 \\ x + 2 &= 3 \\ x &= 1 \end{aligned}$$

$$\therefore (x, y) = (1, 1)$$

Exam­ple 3#

Solve $2x - 3y = 4$ and $5x + 2y = 1$.

$$\begin{aligned} 2x - 3y &= 4 \quad \text{(1)} \\ 5x + 2y &= 1 \quad \text{(2)} \end{aligned}$$

Make the $x$ coef­fi­cients $10$: mul­ti­ply (1) by $5$ and (2) by $2$.

$$\begin{aligned} \text{(1)} \times 5: \quad 10x - 15y &= 20 \\ \text{(2)} \times 2: \quad 10x + 4y &= 2 \end{aligned}$$

Sub­tract: change the lower line to $-10x - 4y = -2$ and add.

$$\displaystyle \begin{aligned} -19y &= 18 \\ y &= -\frac{18}{19} \end{aligned}$$

Sub­sti­tute in (1):

$$\displaystyle \begin{aligned} 2x - 3\left(-\frac{18}{19}\right) &= 4 \\ 2x + \frac{54}{19} &= 4 \\ 2x &= 4 - \frac{54}{19} \\ 2x &= \frac{76 - 54}{19} \\ 2x &= \frac{22}{19} \\ x &= \frac{11}{19} \end{aligned}$$

$$\displaystyle \therefore (x, y) = \left(\frac{11}{19}, -\frac{18}{19}\right)$$

Check in (2): $\displaystyle 5 \times \frac{11}{19} + 2 \times \left(-\frac{18}{19}\right) = \frac{55 - 36}{19} = \frac{19}{19} = 1$.

Exam­ple 4: a word prob­lem#

Three pens and two note­books cost ₹110. Two pens and three note­books cost ₹140. Find the cost of one pen and of one note­book.

Let a pen cost $x$ rupees and a note­book $y$ rupees.

$$\begin{aligned} 3x + 2y &= 110 \quad \text{(1)} \\ 2x + 3y &= 140 \quad \text{(2)} \end{aligned}$$

Elim­i­nate $y$: mul­ti­ply (1) by $3$ and (2) by $2$.

$$\begin{aligned} 9x + 6y &= 330 \\ 4x + 6y &= 280 \end{aligned}$$

Sub­tract: $5x = 50$, so $x = 10$. Then from (1), $30 + 2y = 110$, so $2y = 80$ and $y = 40$. A pen costs ₹10 and a note­book ₹40. Check in (2): $20 + 120 = 140$.

Prac­tice#

Solve each pair by the elim­i­na­tion method.

1. $x + y = 11$ and $x - y = 5$
2. $2x + 3y = 13$ and $2x - y = 1$
3. $3x + 2y = 16$ and $x + 2y = 8$
4. $5x - 2y = 4$ and $3x + 4y = 18$
5. $4x + 3y = 6$ and $3x - 2y = 13$
6. $2x + 5y = 1$ and $3x + 2y = 7$
7. $3x - 4y = 10$ and $4x + 3y = 5$
8. $7x + 3y = 1$ and $5x + 2y = 0$
9. $3x + 4y = 2$ and $5x - 2y = 3$
10. Four chairs and three tables cost ₹6200. Two chairs and five tables cost ₹8000. Find the cost of one chair and of one table.

Answers#