Elimination Method for Pair of Linear Equations, Class 10

Sub­sti­tu­tion always works, but rear­rang­ing an equa­tion to iso­late one vari­able can cre­ate frac­tions before the real work begins. The elim­i­na­tion method, the sec­ond alge­braic method in Chap­ter 3 of the NCERT Class 10 text­book, often avoids this. Instead of rewrit­ing one equa­tion and feed­ing it into the other, you work with both whole equa­tions at once and com­bine them so that one vari­able can­cels out, or is elim­i­nated. What remains is a sim­ple equa­tion in a sin­gle vari­able. Elim­i­na­tion is the quick­est route in many word prob­lems on incomes, prices, dig­its and ages, and the same idea of com­bin­ing equa­tions is used later for larger sys­tems of equa­tions.

What the elim­i­na­tion method is#

The idea in plain words#

If a vari­able, say $y$, has the same coef­fi­cient in both equa­tions, sub­tract­ing one whole equa­tion from the other makes the $y$-terms can­cel exactly. If the coef­fi­cients are equal in size but oppo­site in sign, such as $+3y$ and $-3y$, adding the equa­tions does the same job. Equa­tions rarely come with match­ing coef­fi­cients, so the first move is usu­ally to mul­ti­ply each equa­tion by a suit­able non-zero num­ber to make them match.

Why it works#

Two facts jus­tify every step:

• Mul­ti­ply­ing both sides of an equa­tion by a non-zero num­ber gives an equa­tion with exactly the same solu­tions. For exam­ple, $2x + 3y = 8$ and $4x + 6y = 16$ are sat­is­fied by the same pairs.
• If $(x, y)$ sat­is­fies two equa­tions, it also sat­is­fies their sum and their dif­fer­ence, because equal quan­ti­ties added to (or sub­tracted from) equal quan­ti­ties give equal results.

So any solu­tion of the orig­i­nal pair must sat­isfy the one-vari­able equa­tion obtained after elim­i­na­tion. Sub­sti­tut­ing the value back into an orig­i­nal equa­tion then recov­ers the full solu­tion, and check­ing in both equa­tions con­firms it.

The method in steps#

1. Mul­ti­ply one or both equa­tions by suit­able non-zero num­bers so that one vari­able has numer­i­cally equal coef­fi­cients in the two equa­tions. The low­est com­mon mul­ti­ple of the two coef­fi­cients is usu­ally the best tar­get.
2. Sub­tract the equa­tions if those coef­fi­cients have the same sign; add them if the signs are oppo­site. One vari­able dis­ap­pears.
3. Solve the result­ing equa­tion in one vari­able.
4. Sub­sti­tute this value into either of the orig­i­nal equa­tions to find the other vari­able, and check the pair in both equa­tions.

If in Step 2 both vari­ables dis­ap­pear, the left­over state­ment tells you the type of pair: a true state­ment such as $0 = 0$ means infi­nitely many solu­tions; a false state­ment such as $0 = 9$ means no solu­tion.

Worked exam­ples#

Exam­ple 1: coef­fi­cients already oppo­site#

Solve $x + 2y = 8$ and $3x - 2y = 8$.

The $y$-coef­fi­cients are $+2$ and $-2$, so add the equa­tions:

$$(x + 2y) + (3x - 2y) = 8 + 8 \;\Rightarrow\; 4x = 16 \;\Rightarrow\; x = 4$$

Then from $x + 2y = 8$: $4 + 2y = 8$, so $y = 2$.

Check. $4 + 2(2) = 8$ and $3(4) - 2(2) = 8$. So $x = 4$, $y = 2$.

Exam­ple 2: mul­ti­ply­ing both equa­tions#

Solve $5x + 3y = 21 \quad (1)$ and $2x - 5y = -4 \quad (2)$.

The $y$-coef­fi­cients are $3$ and $-5$; their LCM is $15$. Mul­ti­ply (1) by 5 and (2) by 3:

$$25x + 15y = 105 \quad (3) \qquad 6x - 15y = -12 \quad (4)$$

The signs of the $y$-terms are oppo­site, so add (3) and (4):

$$31x = 93 \;\Rightarrow\; x = 3$$

From (1): $15 + 3y = 21$, so $3y = 6$ and $y = 2$.

Check. $5(3) + 3(2) = 21$ and $2(3) - 5(2) = -4$. So $x = 3$, $y = 2$.

Exam­ple 3: incomes and sav­ings#

The ratio of the incomes of two per­sons is $9 : 7$ and the ratio of their expen­di­tures is $4 : 3$. If each of them saves ₹2000 per month, find their monthly incomes.

Because the incomes are in the ratio $9 : 7$, call them ₹$9x$ and ₹$7x$. Sim­i­larly, call the expen­di­tures ₹$4y$ and ₹$3y$. Since income minus expen­di­ture equals sav­ings,

$$9x - 4y = 2000 \quad (1) \qquad 7x - 3y = 2000 \quad (2)$$

Step 1: match a coef­fi­cient. Mul­ti­ply (1) by 3 and (2) by 4, so that $y$ has coef­fi­cient $-12$ in both:

$$27x - 12y = 6000 \quad (3) \qquad 28x - 12y = 8000 \quad (4)$$

Step 2: sub­tract. The $y$-terms have the same sign, so sub­tract (3) from (4):

$$(28x - 27x) - (12y - 12y) = 8000 - 6000 \;\Rightarrow\; x = 2000$$

Step 3: sub­sti­tute back. Put $x = 2000$ in (1):

$$18000 - 4y = 2000 \;\Rightarrow\; 4y = 16000 \;\Rightarrow\; y = 4000$$

So the incomes are $9x = 18000$ and $7x = 14000$, that is, ₹18,000 and ₹14,000 per month.

Check. $18000 : 14000 = 9 : 7$. The expen­di­tures are $18000 - 2000 = 16000$ and $14000 - 2000 = 12000$, and $16000 : 12000 = 4 : 3$, which also equals $4y : 3y$ with $y = 4000$.

Exam­ple 4: when elim­i­na­tion shows there is no solu­tion#

Solve $2x + 3y = 8 \quad (1)$ and $4x + 6y = 7 \quad (2)$.

Mul­ti­ply (1) by 2 so that both equa­tions con­tain $4x$:

$$4x + 6y = 16 \quad (3)$$

Sub­tract (2) from (3):

$$(4x - 4x) + (6y - 6y) = 16 - 7 \;\Rightarrow\; 0 = 9$$

Both vari­ables have van­ished and the state­ment is false, so the pair has no solu­tion. The ratio test agrees: $\displaystyle \frac{a_1}{a_2} = \frac{2}{4}$ and $\displaystyle \frac{b_1}{b_2} = \frac{3}{6}$ are equal, but $\displaystyle \frac{c_1}{c_2} = \frac{8}{7}$ is dif­fer­ent, which is the pat­tern for par­al­lel lines.

Com­pare this with $3x - y = 2$ and $6x - 2y = 4$. Dou­bling the first and sub­tract­ing gives $0 = 0$, which is always true, so that pair has infi­nitely many solu­tions: the two equa­tions describe the same line.

Exam­ple 5: a puz­zle about dig­its#

The sum of a two-digit num­ber and the num­ber obtained by revers­ing its dig­its is 66. If the dig­its of the num­ber dif­fer by 2, find the num­ber. How many such num­bers are there?

Let the tens digit be $x$ and the units digit be $y$. The num­ber is $10x + y$ (for exam­ple, $56 = 10 \times 5 + 6$), and the reversed num­ber is $10y + x$.

The sum con­di­tion gives

$$(10x + y) + (10y + x) = 66 \;\Rightarrow\; 11(x + y) = 66 \;\Rightarrow\; x + y = 6 \quad (1)$$

"The dig­its dif­fer by 2" can mean $x - y = 2 \quad (2)$ or $y - x = 2 \quad (3)$, since we do not know which digit is larger. Solve both cases.

Case 1: (1) and (2). Adding them elim­i­nates $y$: $2x = 8$, so $x = 4$ and $y = 2$. The num­ber is 42.

Case 2: (1) and (3). Adding them elim­i­nates $x$: $2y = 8$, so $y = 4$ and $x = 2$. The num­ber is 24.

Check. $42 + 24 = 66$, and in each num­ber the dig­its 4 and 2 dif­fer by 2. So there are two such num­bers, 42 and 24.

Choos­ing among the three meth­ods#

You now have three ways to solve a pair of lin­ear equa­tions in two vari­ables:

• Graph­i­cal method: draw both lines and see where they meet. It shows most clearly why a pair can have one solu­tion, no solu­tion or infi­nitely many (inter­sect­ing, par­al­lel or coin­ci­dent lines), but exact val­ues are hard to read unless they are inte­gers.
• Sub­sti­tu­tion method: express one vari­able in terms of the other and sub­sti­tute. It is usu­ally quick­est when some vari­able has coef­fi­cient $1$ or $-1$.
• Elim­i­na­tion method: scale the equa­tions so one vari­able has match­ing coef­fi­cients, then add or sub­tract. It is usu­ally quick­est when the coef­fi­cients are larger num­bers or already line up.

All three meth­ods always give the same answer for the same pair; they are dif­fer­ent routes to one des­ti­na­tion. With prac­tice you will choose the short­est route by look­ing at the coef­fi­cients.

Com­mon mis­takes#

• Mul­ti­ply­ing only one side of an equa­tion. Every term on both sides must be mul­ti­plied, includ­ing the con­stant.
• Adding when you should sub­tract, or the reverse. Same signs: sub­tract. Oppo­site signs: add.
• Sign errors while sub­tract­ing a whole equa­tion. Write it with brack­ets, such as $(28x - 12y) - (27x - 12y)$, before sim­pli­fy­ing.
• Sub­sti­tut­ing back into a scaled equa­tion that was copied wrongly. It is safer to use one of the orig­i­nal equa­tions.
• In digit prob­lems, writ­ing the num­ber as $x + y$ instead of $10x + y$, or for­get­ting that "dif­fer by 2" allows two cases.
• Treat­ing $0 = 0$ as "no solu­tion". A true state­ment means infi­nitely many solu­tions; only a false one, such as $0 = 9$, means none.

Try these#

1. Solve $x + y = 10$ and $x - y = 4$. Answer: $x = 7$, $y = 3$.
2. Solve $3x + 2y = 11$ and $2x + 3y = 4$. Answer: $x = 5$, $y = -2$.
3. Solve $4x - 3y = 1$ and $12x - 9y = 3$. Answer: infi­nitely many solu­tions.
4. Solve $2x - 5y = 3$ and $4x - 10y = 5$. Answer: no solu­tion.
5. The dig­its of a two-digit num­ber add up to 12. Revers­ing the dig­its increases the num­ber by 18. Find the num­ber. Answer: 57.
6. 3 chairs and 2 tables cost ₹4,700, while 5 chairs and 3 tables cost ₹7,400. Find the cost of a chair and of a table. Answer: a chair costs ₹700 and a table ₹1,300.

Key terms#

Elim­i­na­tion method

An alge­braic method that adds or sub­tracts suit­able mul­ti­ples of two equa­tions so that one vari­able can­cels.

Coef­fi­cient

The num­ber mul­ti­ply­ing a vari­able, such as $-4$ in $-4y$.

Equiv­a­lent equa­tions

Equa­tions with exactly the same solu­tions, for exam­ple an equa­tion and a non-zero mul­ti­ple of it.

Con­sis­tent pair

A pair of equa­tions with at least one solu­tion.

Incon­sis­tent pair

A pair with no solu­tion; elim­i­na­tion leaves a false state­ment such as $0 = 9$.

Depen­dent pair

A pair with infi­nitely many solu­tions; elim­i­na­tion leaves a true state­ment such as $0 = 0$.

Place value form

Writ­ing a two-digit num­ber with tens digit $x$ and units digit $y$ as $10x + y$.

Com­mon ques­tions#

How do I decide which vari­able to elim­i­nate?#

Choose the vari­able whose coef­fi­cients need the small­est mul­ti­pli­ers to match, or whose coef­fi­cients already match or are oppo­site. Either choice leads to the same answer.

Can I divide an equa­tion instead of mul­ti­ply­ing?#

Yes. Divid­ing by a non-zero num­ber also gives an equiv­a­lent equa­tion. For exam­ple, $50x + 100y = 2000$ can be divided by 50 to give $x + 2y = 40$.

Why sub­sti­tute into an orig­i­nal equa­tion at the end?#

The orig­i­nal equa­tions are the ones you copied from the ques­tion, so using them avoids car­ry­ing for­ward any slip made while scal­ing.

Is elim­i­na­tion bet­ter than sub­sti­tu­tion?#

Nei­ther is always bet­ter. Elim­i­na­tion is often faster when coef­fi­cients are larger or already match; sub­sti­tu­tion is often faster when a vari­able has coef­fi­cient 1.

How does elim­i­na­tion show the type of pair?#

If one vari­able remains, there is exactly one solu­tion. If both van­ish, a true state­ment means infi­nitely many solu­tions and a false state­ment means no solu­tion.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Hall, H. S. and Knight, S. R. Higher Alge­bra. Macmil­lan.