NCERT Class 10 Exercise 3.3 Solutions: Elimination Method

Ques­tion 1: solve by elim­i­na­tion and by sub­sti­tu­tion#

Solve the fol­low­ing pairs of lin­ear equa­tions by the elim­i­na­tion method and the sub­sti­tu­tion method.

Ques­tion 1 (i)#

$x + y = 5$ and $2x - 3y = 4$

Elim­i­na­tion. The $y$-coef­fi­cients are $1$ and $-3$. Mul­ti­ply the first equa­tion by 3:

$$3x + 3y = 15$$

The $y$-terms now have oppo­site signs, so add this to $2x - 3y = 4$:

$$\displaystyle 5x = 19 \;\Rightarrow\; x = \frac{19}{5}$$

From $x + y = 5$: $\displaystyle y = 5 - \frac{19}{5} = \frac{25 - 19}{5} = \frac{6}{5}$.

Sub­sti­tu­tion. From $x + y = 5$, $y = 5 - x$. Sub­sti­tute into $2x - 3y = 4$:

$$\displaystyle \begin{aligned} 2x - 3(5 - x) &= 4 \\ 2x - 15 + 3x &= 4 \\ 5x &= 19 \\ x &= \frac{19}{5} \end{aligned}$$

Then $\displaystyle y = 5 - \frac{19}{5} = \frac{6}{5}$, the same as before.

Check: $\displaystyle \frac{19}{5} + \frac{6}{5} = \frac{25}{5} = 5$ and $\displaystyle \frac{38}{5} - \frac{18}{5} = \frac{20}{5} = 4$.

Answer: $\displaystyle x = \frac{19}{5}$, $\displaystyle y = \frac{6}{5}$.

Ques­tion 1 (ii)#

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

Elim­i­na­tion. The $y$-coef­fi­cients are $4$ and $-2$. Mul­ti­ply the sec­ond equa­tion by 2:

$$4x - 4y = 4$$

The $y$-terms now have oppo­site signs, so add this to $3x + 4y = 10$:

$$7x = 14 \;\Rightarrow\; x = 2$$

From $2x - 2y = 2$: $4 - 2y = 2$, so $2y = 2$ and $y = 1$.

Sub­sti­tu­tion. Divid­ing $2x - 2y = 2$ by 2 gives $x - y = 1$, so $x = y + 1$. Sub­sti­tute into $3x + 4y = 10$:

$$\begin{aligned} 3(y + 1) + 4y &= 10 \\ 7y + 3 &= 10 \\ 7y &= 7 \\ y &= 1 \end{aligned}$$

Then $x = 1 + 1 = 2$, the same as before.

Check: $3(2) + 4(1) = 10$ and $2(2) - 2(1) = 2$.

Answer: $x = 2$, $y = 1$.

Ques­tion 1 (iii)#

$3x - 5y - 4 = 0$ and $9x = 2y + 7$

First write both in the form $ax + by = c$:

$$3x - 5y = 4 \quad (1) \qquad 9x - 2y = 7 \quad (2)$$

Elim­i­na­tion. Mul­ti­ply (1) by 3 so that $x$ has coef­fi­cient 9 in both:

$$9x - 15y = 12 \quad (3)$$

The $x$-terms have the same sign, so sub­tract (2) from (3):

$$\displaystyle (9x - 15y) - (9x - 2y) = 12 - 7 \;\Rightarrow\; -13y = 5 \;\Rightarrow\; y = -\frac{5}{13}$$

Sub­sti­tute into (1):

$$\displaystyle 3x + \frac{25}{13} = 4 \;\Rightarrow\; 3x = \frac{52 - 25}{13} = \frac{27}{13} \;\Rightarrow\; x = \frac{9}{13}$$

Sub­sti­tu­tion. From (1), $\displaystyle x = \frac{4 + 5y}{3}$. Sub­sti­tute into (2):

$$\displaystyle \begin{aligned} 9 \cdot \frac{4 + 5y}{3} - 2y &= 7 \\ 3(4 + 5y) - 2y &= 7 \\ 12 + 13y &= 7 \\ y &= -\frac{5}{13} \end{aligned}$$

Then $\displaystyle x = \frac{1}{3}\left(4 - \frac{25}{13}\right) = \frac{1}{3} \cdot \frac{27}{13} = \frac{9}{13}$, the same as before.

Check: $\displaystyle 3 \cdot \frac{9}{13} - 5 \cdot \left(-\frac{5}{13}\right) - 4 = \frac{27 + 25}{13} - 4 = 4 - 4 = 0$. Also $\displaystyle 9 \cdot \frac{9}{13} = \frac{81}{13}$ and $\displaystyle 2 \cdot \left(-\frac{5}{13}\right) + 7 = \frac{-10 + 91}{13} = \frac{81}{13}$.

Answer: $\displaystyle x = \frac{9}{13}$, $\displaystyle y = -\frac{5}{13}$.

Ques­tion 1 (iv)#

$\displaystyle \frac{x}{2} + \frac{2y}{3} = -1$ and $\displaystyle x - \frac{y}{3} = 3$

Clear the frac­tions: mul­ti­ply the first equa­tion by 6 and the sec­ond by 3.

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

Elim­i­na­tion. The $x$-coef­fi­cients already match, so sub­tract (2) from (1):

$$(3x + 4y) - (3x - y) = -6 - 9 \;\Rightarrow\; 5y = -15 \;\Rightarrow\; y = -3$$

From (2): $3x + 3 = 9$, so $x = 2$.

Sub­sti­tu­tion. From (2), $y = 3x - 9$. Sub­sti­tute into (1):

$$\begin{aligned} 3x + 4(3x - 9) &= -6 \\ 15x - 36 &= -6 \\ 15x &= 30 \\ x &= 2 \end{aligned}$$

Then $y = 6 - 9 = -3$, the same as before.

Check in the orig­i­nal equa­tions: $\displaystyle \frac{2}{2} + \frac{2(-3)}{3} = 1 - 2 = -1$ and $\displaystyle 2 - \frac{-3}{3} = 2 + 1 = 3$.

Answer: $x = 2$, $y = -3$.

Ques­tion 2: form the pair and solve by elim­i­na­tion#

Form the pair of lin­ear equa­tions in the fol­low­ing prob­lems, and find their solu­tions (if they exist) by the elim­i­na­tion method.

Ques­tion 2 (i)#

If we add 1 to the numer­a­tor and sub­tract 1 from the denom­i­na­tor, a frac­tion reduces to 1. It becomes $\displaystyle \frac{1}{2}$ if we only add 1 to the denom­i­na­tor. What is the frac­tion?

Let the frac­tion be $\displaystyle \frac{x}{y}$. The first con­di­tion gives

$$\displaystyle \frac{x + 1}{y - 1} = 1 \;\Rightarrow\; x + 1 = y - 1 \;\Rightarrow\; x - y = -2 \quad (1)$$

The sec­ond con­di­tion gives

$$\displaystyle \frac{x}{y + 1} = \frac{1}{2} \;\Rightarrow\; 2x = y + 1 \;\Rightarrow\; 2x - y = 1 \quad (2)$$

The $y$-coef­fi­cients match, so sub­tract (1) from (2):

$$(2x - y) - (x - y) = 1 - (-2) \;\Rightarrow\; x = 3$$

From (1): $3 - y = -2$, so $y = 5$.

Check: $\displaystyle \frac{3 + 1}{5 - 1} = \frac{4}{4} = 1$ and $\displaystyle \frac{3}{5 + 1} = \frac{3}{6} = \frac{1}{2}$.

Answer: the frac­tion is $\displaystyle \frac{3}{5}$.

Ques­tion 2 (ii)#

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Let Nuri's present age be $x$ years and Sonu's be $y$ years.

Five years ago:

$$x - 5 = 3(y - 5) \;\Rightarrow\; x - 3y = -10 \quad (1)$$

Ten years later:

$$x + 10 = 2(y + 10) \;\Rightarrow\; x - 2y = 10 \quad (2)$$

The $x$-coef­fi­cients match, so sub­tract (1) from (2):

$$(x - 2y) - (x - 3y) = 10 - (-10) \;\Rightarrow\; y = 20$$

From (2): $x - 40 = 10$, so $x = 50$.

Check against the words: five years ago Nuri was 45 and Sonu 15, and $45 = 3 \times 15$. Ten years later Nuri will be 60 and Sonu 30, and $60 = 2 \times 30$.

Answer: Nuri is 50 years old and Sonu is 20 years old.

Ques­tion 2 (iii)#

The sum of the dig­its of a two-digit num­ber is 9. Also, nine times this num­ber is twice the num­ber obtained by revers­ing the order of the dig­its. Find the num­ber.

Let the tens digit be $x$ and the units digit be $y$. The num­ber is $10x + y$ and the reversed num­ber is $10y + x$.

$$x + y = 9 \quad (1)$$

$$9(10x + y) = 2(10y + x) \;\Rightarrow\; 90x + 9y = 20y + 2x \;\Rightarrow\; 88x - 11y = 0 \;\Rightarrow\; 8x - y = 0 \quad (2)$$

The $y$-terms have oppo­site signs, so add (1) and (2):

$$9x = 9 \;\Rightarrow\; x = 1$$

From (1): $y = 9 - 1 = 8$.

Check: the num­ber is 18; $1 + 8 = 9$; $9 \times 18 = 162$ and $2 \times 81 = 162$.

Answer: the num­ber is 18.

Ques­tion 2 (iv)#

Meena went to a bank to with­draw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.

Let $x$ be the num­ber of ₹50 notes and $y$ the num­ber of ₹100 notes.

$$x + y = 25 \quad (1)$$

$$50x + 100y = 2000 \;\Rightarrow\; x + 2y = 40 \quad (2)$$

(Equa­tion (2) was divided by 50.) The $x$-coef­fi­cients match, so sub­tract (1) from (2):

$$(x + 2y) - (x + y) = 40 - 25 \;\Rightarrow\; y = 15$$

From (1): $x = 25 - 15 = 10$.

Check: $10 + 15 = 25$ notes, and $10 \times 50 + 15 \times 100 = 500 + 1500 = 2000$.

Answer: Meena received 10 notes of ₹50 and 15 notes of ₹100.

Ques­tion 2 (v)#

A lend­ing library has a fixed charge for the first three days and an addi­tional charge for each day there­after. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Let the fixed charge be ₹$x$ and the charge for each extra day be ₹$y$. Seven days means $7 - 3 = 4$ extra days, and five days means $5 - 3 = 2$ extra days. So

$$x + 4y = 27 \quad (1) \qquad x + 2y = 21 \quad (2)$$

Sub­tract (2) from (1):

$$2y = 6 \;\Rightarrow\; y = 3$$

From (2): $x = 21 - 6 = 15$.

Check: Saritha pays $15 + 4 \times 3 = 27$ and Susy pays $15 + 2 \times 3 = 21$.

Answer: the fixed charge is ₹15 and the charge for each extra day is ₹3.

Why this works: the fixed charge already cov­ers the first three days, so only the days beyond three are mul­ti­plied by $y$. Writ­ing $x + 7y$ would be a com­mon mis­take.

Across Ques­tion 2, the alge­bra is short once the equa­tions are right. The real skill is trans­lat­ing each sen­tence exactly, so read every ques­tion twice before writ­ing an equa­tion.