Substitution Method for Pair of Linear Equations, Class 10

The idea in plain words#

Use one equa­tion to write one vari­able, say $y$, entirely in terms of the other vari­able, $x$. Then, wher­ever $y$ appears in the other equa­tion, put in that expres­sion instead. The sec­ond equa­tion now con­tains only $x$, so it can be solved like any ordi­nary equa­tion in one vari­able. Once $x$ is known, go back and find $y$.

Why it works#

If $(x, y)$ sat­is­fies both equa­tions, then in par­tic­u­lar it sat­is­fies the rearranged first equa­tion, say $y = mx + k$. Replac­ing $y$ by $mx + k$ in the sec­ond equa­tion there­fore does not change which val­ues of $x$ are pos­si­ble: any solu­tion of the pair must sat­isfy the new one-vari­able equa­tion. Con­versely, if $x$ sat­is­fies the new equa­tion and we define $y = mx + k$, then both orig­i­nal equa­tions hold. So the one-vari­able equa­tion car­ries exactly the same infor­ma­tion as the pair. Noth­ing is lost and noth­ing extra is intro­duced, which is why the method is reli­able.

The method in steps#

1. From either equa­tion, express one vari­able in terms of the other. Choose the vari­able whose coef­fi­cient is $1$ or $-1$ if you can, because this avoids frac­tions.
2. Sub­sti­tute this expres­sion into the other equa­tion. You now have one equa­tion in one vari­able. Solve it. (If the vari­able can­cels com­pletely, read the spe­cial cases below.)
3. Sub­sti­tute the value just found into the expres­sion from Step 1 to get the sec­ond vari­able.
4. Check the pair in both orig­i­nal equa­tions.

Worked exam­ples#

Exam­ple 1: a warm-up#

Solve $2x + y = 11$ and $x - y = 1$.

Step 1. In the first equa­tion $y$ has coef­fi­cient $1$, so $y = 11 - 2x$.

Step 2. Sub­sti­tute into the sec­ond equa­tion:

$$\begin{aligned} x - (11 - 2x) &= 1 \\ 3x - 11 &= 1 \\ 3x &= 12 \\ x &= 4 \end{aligned}$$

Step 3. $y = 11 - 2(4) = 3$.

Check. $2(4) + 3 = 11$ and $4 - 3 = 1$. Both hold, so $x = 4$, $y = 3$.

Exam­ple 2: an answer no graph could show#

Solve $7x - 15y = 2 \quad (1)$ and $x + 2y = 3 \quad (2)$.

Step 1. Equa­tion (2) is eas­ier to rearrange because $x$ has coef­fi­cient $1$:

$$x = 3 - 2y \quad (3)$$

Step 2. Replace $x$ in equa­tion (1) by $3 - 2y$:

$$\displaystyle \begin{aligned} 7(3 - 2y) - 15y &= 2 \\ 21 - 14y - 15y &= 2 \\ -29y &= -19 \\ y &= \frac{19}{29} \end{aligned}$$

Step 3. Put this value into equa­tion (3):

$$\displaystyle x = 3 - 2\left(\frac{19}{29}\right) = \frac{87 - 38}{29} = \frac{49}{29}$$

Check. $\displaystyle 7 \cdot \frac{49}{29} - 15 \cdot \frac{19}{29} = \frac{343 - 285}{29} = \frac{58}{29} = 2$, and $\displaystyle \frac{49}{29} + \frac{38}{29} = \frac{87}{29} = 3$. So $\displaystyle x = \frac{49}{29}$, $\displaystyle y = \frac{19}{29}$.

Exam­ple 3: Aftab and his daugh­ter#

Aftab tells his daugh­ter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." Find their present ages.

Let Aftab's present age be $s$ years and his daugh­ter's be $t$ years.

Seven years ago their ages were $s - 7$ and $t - 7$, so

$$s - 7 = 7(t - 7) \;\Rightarrow\; s - 7t + 42 = 0 \quad (1)$$

Three years from now their ages will be $s + 3$ and $t + 3$, so

$$s + 3 = 3(t + 3) \;\Rightarrow\; s - 3t = 6 \quad (2)$$

From (2), $s = 3t + 6$. Sub­sti­tute into (1):

$$\begin{aligned} (3t + 6) - 7t + 42 &= 0 \\ -4t + 48 &= 0 \\ t &= 12 \end{aligned}$$

Then $s = 3(12) + 6 = 42$.

Check against the words. Seven years ago Aftab was $35$ and his daugh­ter $5$, and $35 = 7 \times 5$. Three years from now they will be $45$ and $15$, and $45 = 3 \times 15$. So Aftab is 42 years old and his daugh­ter is 12.

Exam­ple 4: when every value works#

In a shop, 2 pen­cils and 3 erasers cost ₹9, and 4 pen­cils and 6 erasers cost ₹18. Find the cost of one pen­cil and one eraser.

Let a pen­cil cost ₹$x$ and an eraser ₹$y$. Then

$$2x + 3y = 9 \quad (1) \qquad 4x + 6y = 18 \quad (2)$$

From (1), $\displaystyle x = \frac{9 - 3y}{2}$. Sub­sti­tute into (2):

$$\displaystyle 4\left(\frac{9 - 3y}{2}\right) + 6y = 18 \;\Rightarrow\; 18 - 6y + 6y = 18 \;\Rightarrow\; 18 = 18$$

The vari­able has dis­ap­peared and what is left is always true. Every value of $y$ works, with $\displaystyle x = \frac{9 - 3y}{2}$. The rea­son is that equa­tion (2) is just equa­tion (1) mul­ti­plied by 2, so the shop has really given only one piece of infor­ma­tion. The pair has infi­nitely many solu­tions, and the costs can­not be pinned down from these two state­ments alone.

Exam­ple 5: when noth­ing works#

Two rails are rep­re­sented by $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$. Will the rails cross?

From the first equa­tion, $x = 4 - 2y$. Sub­sti­tute into the sec­ond:

$$2(4 - 2y) + 4y - 12 = 0 \;\Rightarrow\; 8 - 4y + 4y - 12 = 0 \;\Rightarrow\; -4 = 0$$

This state­ment is false what­ever $y$ is. So no pair $(x, y)$ sat­is­fies both equa­tions: the pair has no solu­tion, and the rails never meet. That is exactly what rail­way tracks are built to do; they are par­al­lel.

Read­ing the result: three pos­si­ble out­comes#

After sub­sti­tu­tion, the one-vari­able equa­tion always falls into one of three types, and each type matches a pic­ture of the two lines.

• A nor­mal equa­tion, such as $-29y = -19$: there is exactly one solu­tion, and the lines inter­sect at one point. The pair is con­sis­tent.
• A state­ment that is always true, such as $18 = 18$: the two equa­tions describe the same line, the lines coin­cide, and there are infi­nitely many solu­tions. The pair is depen­dent (and con­sis­tent).
• A state­ment that is always false, such as $-4 = 0$: the lines are par­al­lel and there is no solu­tion. The pair is incon­sis­tent.

You can pre­dict the out­come before solv­ing by com­par­ing the ratios of the coef­fi­cients:

• $\displaystyle \frac{a_1}{a_2} \ne \frac{b_1}{b_2}$: one solu­tion (inter­sect­ing lines).
• $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$: infi­nitely many solu­tions (coin­ci­dent lines).
• $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$: no solu­tion (par­al­lel lines).

For the pen­cils, $\displaystyle \frac{2}{4} = \frac{3}{6} = \frac{-9}{-18}$, so the lines coin­cide. For the rails, $\displaystyle \frac{1}{2} = \frac{2}{4}$ but $\displaystyle \frac{-4}{-12} = \frac{1}{3} \ne \frac{1}{2}$, so the lines are par­al­lel. Sub­sti­tu­tion and the ratio test always agree.

Com­mon mis­takes#

• Sub­sti­tut­ing the expres­sion back into the same equa­tion it came from. This always gives some­thing like $0 = 0$ and tells you noth­ing. Sub­sti­tute into the other equa­tion.
• For­get­ting brack­ets. $x - (14 - x)$ is $2x - 14$, not $-14$; write the bracket first, then expand.
• Sign slips when mov­ing terms: from $x + 2y = 3$ we get $x = 3 - 2y$, not $x = 3 + 2y$.
• Stop­ping after find­ing one vari­able. The answer to a pair is always an ordered pair; find both val­ues.
• Read­ing $18 = 18$ as "no solu­tion" or $-4 = 0$ as "infi­nitely many". A true state­ment means infi­nitely many solu­tions; a false one means none.
• Skip­ping the check. Put both val­ues into both orig­i­nal equa­tions, and in word prob­lems check against the sen­tences, not only against your own equa­tions.

Try these#

1. Solve $x + y = 7$ and $3x - 2y = 11$. Answer: $x = 5$, $y = 2$.
2. Solve $2x - y = 1$ and $3x + 2y = 12$. Answer: $x = 2$, $y = 3$.
3. Solve $x - 3y = 2$ and $2x - 6y = 4$. Answer: infi­nitely many solu­tions, $x = 3y + 2$ for any $y$.
4. Solve $3x + y = 5$ and $6x + 2y = 7$. Answer: no solu­tion (par­al­lel lines).
5. The sum of two num­bers is 50 and one exceeds the other by 14. Find the num­bers. Answer: 32 and 18.
6. 5 pens and 2 note­books cost ₹110, while 3 pens and 4 note­books cost ₹136. Find the cost of each. Answer: a pen costs ₹12 and a note­book ₹25.

Key terms#

Pair of lin­ear equa­tions in two vari­ables

Two equa­tions of the form $ax + by + c = 0$ in the same two vari­ables, con­sid­ered together.

Solu­tion of a pair

An ordered pair $(x, y)$ that sat­is­fies both equa­tions; on a graph, a point com­mon to both lines.

Sub­sti­tu­tion method

An alge­braic method that expresses one vari­able in terms of the other from one equa­tion and puts that expres­sion into the other equa­tion.

Con­sis­tent pair

A pair with at least one solu­tion: either exactly one (inter­sect­ing lines) or infi­nitely many (coin­ci­dent lines).

Incon­sis­tent pair

A pair with no solu­tion; its lines are par­al­lel.

Depen­dent pair

A con­sis­tent pair with infi­nitely many solu­tions, where one equa­tion is a mul­ti­ple of the other.

Coef­fi­cient

The num­ber mul­ti­ply­ing a vari­able, such as $7$ in $7x$.

Com­mon ques­tions#

Which vari­able should I express first?#

Pick a vari­able whose coef­fi­cient is $1$ or $-1$. Then the expres­sion has no frac­tions and the arith­metic stays sim­ple. Any choice gives the same final answer.

Is sub­sti­tu­tion bet­ter than the graph­i­cal method?#

For find­ing exact val­ues, yes. A graph shows how many solu­tions there are and roughly where they lie, but sub­sti­tu­tion gives exact answers even when they are frac­tions or surds.

What does it mean when the vari­able dis­ap­pears?#

If what remains is true, such as $18 = 18$, the equa­tions describe the same line and there are infi­nitely many solu­tions. If it is false, such as $-4 = 0$, the lines are par­al­lel and there is no solu­tion.

Do I have to write the check in the exam?#

It is not always required for marks, but it takes only a line and catches most sign errors. In word prob­lems, check­ing against the orig­i­nal sen­tences is the surest test.

Can sub­sti­tu­tion be used when the equa­tions have dec­i­mals or frac­tions?#

Yes. It usu­ally helps to mul­ti­ply each equa­tion by a suit­able num­ber first so that all coef­fi­cients become whole num­bers.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.