NCERT Class 10 Exercise 3.1 Solutions: Graphical Method

Form­ing equa­tions and solv­ing graph­i­cally#

Ques­tion 1 (i)#

Form the pair of lin­ear equa­tions and find the solu­tion graph­i­cally: 10 stu­dents of Class X took part in a Math­e­mat­ics quiz. If the num­ber of girls is 4 more than the num­ber of boys, find the num­ber of boys and girls who took part in the quiz.

Let the num­ber of boys be $x$ and the num­ber of girls be $y$.

$$x + y = 10 \qquad \text{and} \qquad y - x = 4$$

Plot­ting these points and draw­ing the lines, they meet at $(3, 7)$.

Check: $3 + 7 = 10$ and $7 - 3 = 4$.

Answer: 3 boys and 7 girls took part in the quiz.

Ques­tion 1 (ii)#

5 pen­cils and 7 pens together cost ₹50, whereas 7 pen­cils and 5 pens together cost ₹46. Find the cost of one pen­cil and that of one pen.

Let one pen­cil cost ₹$x$ and one pen cost ₹$y$ (so $x$ and $y$ are num­bers of rupees).

$$5x + 7y = 50 \qquad \text{(1)}$$

$$7x + 5y = 46 \qquad \text{(2)}$$

Points on line (1): $(10, 0)$, $(3, 5)$ and $(-4, 10)$. Points on line (2): $(3, 5)$, $(8, -2)$ and $(-2, 12)$. The $x$-inter­cept of line (2) is $\displaystyle \frac{46}{7} \approx 6.57$, which is not a whole num­ber, so it is bet­ter to plot points with whole-num­ber coor­di­nates.

Both lines pass through $(3, 5)$, so that is where they meet.

Alge­braic con­fir­ma­tion: mul­ti­ply (1) by 7 and (2) by 5 to get $35x + 49y = 350$ and $35x + 25y = 230$. Sub­tract­ing, $24y = 120$, so $y = 5$. Then $5x + 35 = 50$, so $x = 3$.

Check: $5(3) + 7(5) = 15 + 35 = 50$ and $7(3) + 5(5) = 21 + 25 = 46$.

Answer: one pen­cil costs ₹3 and one pen costs ₹5.

Inter­sect­ing, par­al­lel or coin­ci­dent#

Ques­tion 2 (i)#

On com­par­ing the ratios $\displaystyle \frac{a_1}{a_2}$, $\displaystyle \frac{b_1}{b_2}$ and $\displaystyle \frac{c_1}{c_2}$, find out whether the lines inter­sect at a point, are par­al­lel or coin­ci­dent: (i) $5x - 4y + 8 = 0$ and $7x + 6y - 9 = 0$.

Here $a_1 = 5$, $b_1 = -4$, $c_1 = 8$ and $a_2 = 7$, $b_2 = 6$, $c_2 = -9$.

$$\displaystyle \frac{a_1}{a_2} = \frac{5}{7}, \qquad \frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}$$

Since $\displaystyle \frac{5}{7} \neq -\frac{2}{3}$, the lines inter­sect. (One ratio is pos­i­tive and the other neg­a­tive, so they can­not be equal.)

Answer: the lines inter­sect at a point.

Ques­tion 2 (ii)#

(ii) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$

$$\displaystyle \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$$

All three ratios are equal. Indeed, $2(9x + 3y + 12) = 18x + 6y + 24$, so the sec­ond equa­tion is just twice the first.

Answer: the lines are coin­ci­dent.

Ques­tion 2 (iii)#

(iii) $6x - 3y + 10 = 0$ and $2x - y + 9 = 0$

$$\displaystyle \frac{a_1}{a_2} = \frac{6}{2} = 3, \qquad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \qquad \frac{c_1}{c_2} = \frac{10}{9}$$

The first two ratios are equal but the third is dif­fer­ent. Check: $3(2x - y + 9) = 6x - 3y + 27$, which has the same $x$ and $y$ terms as $6x - 3y + 10$ but a dif­fer­ent con­stant, so the two can never both be zero.

Answer: the lines are par­al­lel.

Con­sis­tent or incon­sis­tent#

Ques­tion 3 (i)#

On com­par­ing the ratios, find out whether the fol­low­ing pairs of lin­ear equa­tions are con­sis­tent or incon­sis­tent: (i) $3x + 2y = 5$ and $2x - 3y = 7$.

In gen­eral form: $3x + 2y - 5 = 0$ and $2x - 3y - 7 = 0$.

$$\displaystyle \frac{a_1}{a_2} = \frac{3}{2}, \qquad \frac{b_1}{b_2} = \frac{2}{-3} = -\frac{2}{3}$$

The ratios dif­fer, so the lines inter­sect at one point.

Con­fir­ma­tion: mul­ti­ply­ing the equa­tions by 3 and 2 gives $9x + 6y = 15$ and $4x - 6y = 14$. Adding, $13x = 29$, so $\displaystyle x = \frac{29}{13}$, and then $\displaystyle y = \frac{5 - 3x}{2} = -\frac{11}{13}$. A sin­gle def­i­nite solu­tion exists.

Answer: con­sis­tent (unique solu­tion).

Ques­tion 3 (ii)#

(ii) $2x - 3y = 8$ and $4x - 6y = 9$

In gen­eral form: $2x - 3y - 8 = 0$ and $4x - 6y - 9 = 0$.

$$\displaystyle \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$$

The first two ratios are equal and the third is not. Check: dou­bling the first equa­tion gives $4x - 6y = 16$, but the sec­ond says $4x - 6y = 9$; both can­not be true.

Answer: incon­sis­tent (no solu­tion).

Ques­tion 3 (iii)#

(iii) $\displaystyle \frac{3}{2}x + \frac{5}{3}y = 7$ and $9x - 10y = 14$

In gen­eral form: $\displaystyle \frac{3}{2}x + \frac{5}{3}y - 7 = 0$ and $9x - 10y - 14 = 0$.

$$\displaystyle \frac{a_1}{a_2} = \frac{3/2}{9} = \frac{1}{6}, \qquad \frac{b_1}{b_2} = \frac{5/3}{-10} = -\frac{1}{6}$$

The ratios dif­fer, so the lines inter­sect.

Con­fir­ma­tion: mul­ti­ply the first equa­tion by 6 to get $9x + 10y = 42$. Adding this to $9x - 10y = 14$ gives $18x = 56$, so $\displaystyle x = \frac{28}{9}$; sub­tract­ing gives $20y = 28$, so $\displaystyle y = \frac{7}{5}$. Check in the orig­i­nal first equa­tion: $\displaystyle \frac{3}{2} \cdot \frac{28}{9} + \frac{5}{3} \cdot \frac{7}{5} = \frac{14}{3} + \frac{7}{3} = 7$.

Answer: con­sis­tent (unique solu­tion).

Ques­tion 3 (iv)#

(iv) $5x - 3y = 11$ and $-10x + 6y = -22$

Watch the signs. In gen­eral form: $5x - 3y - 11 = 0$ and $-10x + 6y + 22 = 0$. So $c_1 = -11$ and $c_2 = +22$.

$$\displaystyle \frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}$$

All three ratios are equal. Indeed, $-2(5x - 3y - 11) = -10x + 6y + 22$.

Answer: con­sis­tent, with infi­nitely many solu­tions (coin­ci­dent lines).

Ques­tion 3 (v)#

(v) $\displaystyle \frac{4}{3}x + 2y = 8$ and $2x + 3y = 12$

In gen­eral form: $\displaystyle \frac{4}{3}x + 2y - 8 = 0$ and $2x + 3y - 12 = 0$.

$$\displaystyle \frac{a_1}{a_2} = \frac{4/3}{2} = \frac{2}{3}, \qquad \frac{b_1}{b_2} = \frac{2}{3}, \qquad \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3}$$

All three ratios are equal. Indeed, $\displaystyle \frac{2}{3}(2x + 3y - 12) = \frac{4}{3}x + 2y - 8$.

Answer: con­sis­tent, with infi­nitely many solu­tions (coin­ci­dent lines).

Con­sis­tency and graph­i­cal solu­tions#

Ques­tion 4 (i)#

Which of the fol­low­ing pairs of lin­ear equa­tions are con­sis­tent or incon­sis­tent? If con­sis­tent, obtain the solu­tion graph­i­cally: (i) $x + y = 5$ and $2x + 2y = 10$.

In gen­eral form: $x + y - 5 = 0$ and $2x + 2y - 10 = 0$.

$$\displaystyle \frac{a_1}{a_2} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}$$

All equal, so the lines coin­cide and the pair is con­sis­tent. Plot­ting $x + y = 5$ from the points $(0, 5)$, $(5, 0)$ and $(2, 3)$ gives one line, and $2x + 2y = 10$ passes through exactly the same points (for exam­ple, $2(2) + 2(3) = 10$). Every point of this line is a solu­tion.

Answer: con­sis­tent, with infi­nitely many solu­tions: $y = 5 - x$ for any value of $x$, such as $(0, 5)$, $(2, 3)$ and $(5, 0)$.

Ques­tion 4 (ii)#

(ii) $x - y = 8$ and $3x - 3y = 16$

In gen­eral form: $x - y - 8 = 0$ and $3x - 3y - 16 = 0$.

$$\displaystyle \frac{a_1}{a_2} = \frac{1}{3}, \qquad \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}, \qquad \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$$

The first two ratios are equal and the third is not, so the lines are par­al­lel. Check: tripling the first gives $3x - 3y = 24$, but the sec­ond says $3x - 3y = 16$.

Answer: incon­sis­tent; there is no solu­tion to draw.

Ques­tion 4 (iii)#

(iii) $2x + y - 6 = 0$ and $4x - 2y - 4 = 0$

$$\displaystyle \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$$

The ratios dif­fer, so the pair is con­sis­tent with a unique solu­tion.

Points on $2x + y = 6$: $(0, 6)$, $(3, 0)$, $(2, 2)$. Points on $4x - 2y = 4$: $(0, -2)$, $(1, 0)$, $(2, 2)$. The lines meet at $(2, 2)$.

Alge­braic con­fir­ma­tion: from the first equa­tion $y = 6 - 2x$; then $4x - 2(6 - 2x) - 4 = 0$ gives $8x = 16$, so $x = 2$ and $y = 2$.

Check: $2(2) + 2 - 6 = 0$ and $4(2) - 2(2) - 4 = 0$.

Answer: con­sis­tent; the solu­tion is $x = 2$, $y = 2$.

Ques­tion 4 (iv)#

(iv) $2x - 2y - 2 = 0$ and $4x - 4y - 5 = 0$

$$\displaystyle \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$$

The first two ratios are equal but $\displaystyle \frac{1}{2} \neq \frac{2}{5}$, so the lines are par­al­lel. Check: dou­bling the first gives $4x - 4y = 4$, while the sec­ond says $4x - 4y = 5$.

Answer: incon­sis­tent; there is no solu­tion.

A prob­lem on a rec­tan­gle#

Ques­tion 5#

Half the perime­ter of a rec­tan­gu­lar gar­den, whose length is 4 m more than its width, is 36 m. Find the dimen­sions of the gar­den.

Let the width be $x$ m and the length be $y$ m. Half the perime­ter of a rec­tan­gle is length + width, so

$$x + y = 36 \qquad \text{and} \qquad y - x = 4$$

Adding the two equa­tions: $2y = 40$, so $y = 20$. Then $x = 36 - 20 = 16$. (On a graph, the line $x + y = 36$ through $(0, 36)$ and $(36, 0)$ meets the line $y - x = 4$ through $(0, 4)$ and $(10, 14)$ at $(16, 20)$.)

Check: $20 - 16 = 4$, and half the perime­ter is $20 + 16 = 36$ m.

Answer: the gar­den is 20 m long and 16 m wide.

Build­ing a sec­ond equa­tion#

Ques­tion 6 (i)#

Given the lin­ear equa­tion $2x + 3y - 8 = 0$, write another lin­ear equa­tion in two vari­ables such that the geo­met­ri­cal rep­re­sen­ta­tion of the pair so formed is: (i) inter­sect­ing lines.

We need $\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, so choose coef­fi­cients not in the ratio $2 : 3$. Take

$$x - y - 1 = 0$$

Then $\displaystyle \frac{a_1}{a_2} = \frac{2}{1} = 2$ and $\displaystyle \frac{b_1}{b_2} = \frac{3}{-1} = -3$, which are dif­fer­ent.

Answer: $x - y - 1 = 0$ (many other answers are pos­si­ble).

Ques­tion 6 (ii)#

(ii) par­al­lel lines

We need $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$: keep the $x$ and $y$ coef­fi­cients in the ratio $2 : 3$ and change the con­stant. Take

$$2x + 3y - 20 = 0$$

Then $\displaystyle \frac{a_1}{a_2} = 1$, $\displaystyle \frac{b_1}{b_2} = 1$ and $\displaystyle \frac{c_1}{c_2} = \frac{-8}{-20} = \frac{2}{5} \neq 1$.

Answer: $2x + 3y - 20 = 0$ (any $2x + 3y + c = 0$ with $c \neq -8$, or a mul­ti­ple of such an equa­tion, also works).

Ques­tion 6 (iii)#

(iii) coin­ci­dent lines

We need all three ratios equal, so mul­ti­ply the whole equa­tion by a non-zero con­stant. Tak­ing 2,

$$4x + 6y - 16 = 0$$

Then $\displaystyle \frac{a_1}{a_2} = \frac{2}{4}$, $\displaystyle \frac{b_1}{b_2} = \frac{3}{6}$ and $\displaystyle \frac{c_1}{c_2} = \frac{-8}{-16}$, all equal to $\displaystyle \frac{1}{2}$.

Answer: $4x + 6y - 16 = 0$ (any non-zero mul­ti­ple of the given equa­tion works).

A tri­an­gle from two lines#

Ques­tion 7#

Draw the graphs of the equa­tions $x - y + 1 = 0$ and $3x + 2y - 12 = 0$. Deter­mine the coor­di­nates of the ver­tices of the tri­an­gle formed by these lines and the $x$-axis, and shade the tri­an­gu­lar region.

Points on $x - y + 1 = 0$ (that is, $y = x + 1$): $(-1, 0)$, $(0, 1)$, $(2, 3)$.

Points on $3x + 2y - 12 = 0$: $(4, 0)$, $(0, 6)$, $(2, 3)$.

The three ver­tices of the tri­an­gle are:

• where $x - y + 1 = 0$ meets the $x$-axis ($y = 0$): $x = -1$, giv­ing $A(-1, 0)$;
• where $3x + 2y - 12 = 0$ meets the $x$-axis: $3x = 12$, $x = 4$, giv­ing $B(4, 0)$;
• where the two lines meet: sub­sti­tut­ing $y = x + 1$ into the sec­ond equa­tion, $3x + 2x + 2 - 12 = 0$, so $5x = 10$, $x = 2$ and $y = 3$, giv­ing $C(2, 3)$.

Check: $-1 - 0 + 1 = 0$; $3(4) + 0 - 12 = 0$; $2 - 3 + 1 = 0$ and $6 + 6 - 12 = 0$.

As an extra check, the base $AB$ is $4 - (-1) = 5$ units and the height is 3 units, so the area is $\displaystyle \frac{1}{2} \times 5 \times 3 = 7.5$ square units.

Answer: the ver­tices are $(-1, 0)$, $(4, 0)$ and $(2, 3)$; the region enclosed by the two lines and the $x$-axis is shaded.

Key terms#

Pair of lin­ear equa­tions in two vari­ables

Two equa­tions of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ con­sid­ered together.

Con­sis­tent pair

A pair with at least one solu­tion; its lines inter­sect or coin­cide.

Incon­sis­tent pair

A pair with no solu­tion; its lines are par­al­lel.

Depen­dent pair

A pair whose equa­tions rep­re­sent the same line, so it has infi­nitely many solu­tions.

Inter­sect­ing lines

Lines with exactly one com­mon point, which gives the unique solu­tion.

Par­al­lel lines

Lines in a plane that never meet.

Coin­ci­dent lines

Lines that lie on top of each other and share every point.

Inter­cept on the $x$-axis

The point where a line meets the $x$-axis, found by putting $y = 0$.

Com­mon ques­tions#

Do I have to draw a graph if the ques­tion asks for a graph­i­cal solu­tion?#

Yes. In the exam, draw the lines on graph paper from a table of points and mark the inter­sec­tion. You may con­firm the answer alge­braically, but the graph is what the ques­tion asks for.

Why does $-10x + 6y = -22$ have $c_2 = 22$?#

Mov­ing $-22$ to the left-hand side changes its sign, giv­ing $-10x + 6y + 22 = 0$. Using $c_2 = -22$ would give the ratio $\displaystyle \frac{1}{2}$ and a wrong answer.

Is there only one cor­rect answer to Ques­tion 6?#

No. Any equa­tion meet­ing the required ratio con­di­tion is cor­rect. Always show the ratio check with the equa­tion you choose.

What should I write when a pair is coin­ci­dent and the ques­tion says "obtain the solu­tion"?#

State that there are infi­nitely many solu­tions, write the gen­eral form (for Ques­tion 4 (i), $y = 5 - x$) and list a few exam­ples.

How do I choose points when the inter­cepts are frac­tions?#

Try small whole val­ues of $x$ until $y$ comes out whole. For $7x + 5y = 46$, $x = 3$ gives $y = 5$ and $x = 8$ gives $y = -2$.

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.