Graphical Method for Pair of Linear Equations, Class 10

Akhila went to a fair in her vil­lage. She wanted to ride the Giant Wheel and play Hoopla, a game in which you throw a ring over the items kept in a stall and win the item if the ring cov­ers it com­pletely. She played Hoopla half as many times as she rode the Giant Wheel. Each ride cost ₹3 and each game of Hoopla cost ₹4. She spent ₹20 in all. How many rides did she take, and how many times did she play Hoopla?

You could guess and check, but there is a neater way. Two unknown num­bers and two facts about them give a pair of lin­ear equa­tions in two vari­ables. This les­son, from Chap­ter 3 of the NCERT Class 10 text­book, shows how to solve such a pair by draw­ing both lines on one graph, and how to pre­dict from the coef­fi­cients alone whether a solu­tion exists. The same ideas are used in money prob­lems, mix­ture and speed prob­lems, and any sit­u­a­tion where two con­di­tions must hold together.

What a pair of lin­ear equa­tions is

Lin­ear equa­tion in two vari­ables

A lin­ear equa­tion in two vari­ables $x$ and $y$ can be writ­ten as

$$ax + by + c = 0$$

where $a$, $b$, $c$ are real num­bers and $a$ and $b$ are not both zero. Each vari­able appears only to the first power, and $x$ and $y$ are never mul­ti­plied together.

Set­ting up Akhi­la's pair

Let $x$ be the num­ber of rides on the Giant Wheel and $y$ the num­ber of Hoopla games. "Hoopla was played half as many times" gives $\displaystyle y = \dfrac{x}{2}$, that is,

$$x - 2y = 0$$

The spend­ing (₹3 per ride, ₹4 per game, ₹20 in total) gives

$$3x + 4y = 20$$

A solu­tion of the pair is a pair of val­ues of $x$ and $y$ that makes both equa­tions true at the same time.

Turn­ing each equa­tion into a line

The key fact

Every solu­tion $(x, y)$ of a lin­ear equa­tion in two vari­ables is a point on a straight line, and every point on that line is a solu­tion. So a pair of lin­ear equa­tions is a pair of straight lines on the same axes, and solv­ing the pair means find­ing the point or points that lie on both lines.

The three pos­si­ble pic­tures

Two lines in a plane can behave in only three ways:

1. They inter­sect at exactly one point. The pair has a unique solu­tion. Such a pair is called con­sis­tent.
2. They are par­al­lel. They never meet, so the pair has no solu­tion. Such a pair is called incon­sis­tent.
3. They coin­cide (they are the same line). Every point on the line is a solu­tion, so there are infi­nitely many solu­tions. Such a pair is called depen­dent, and a depen­dent pair is also con­sis­tent.

Pre­dict­ing the pic­ture from the coef­fi­cients

The ratio test

Write the pair in gen­eral form:

$$a_1x + b_1y + c_1 = 0 \qquad \text{and} \qquad a_2x + b_2y + c_2 = 0$$

Then com­pare the ratios of the coef­fi­cients:

• If $\displaystyle \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the lines inter­sect: a unique solu­tion (con­sis­tent).
• If $\displaystyle \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the lines coin­cide: infi­nitely many solu­tions (depen­dent and con­sis­tent).
• If $\displaystyle \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the lines are par­al­lel: no solu­tion (incon­sis­tent).

Why the test works

If $\displaystyle \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = k$, then $a_1 = ka_2$ and $b_1 = kb_2$, so the first equa­tion is $k(a_2x + b_2y) + c_1 = 0$. Both lines then have the same steep­ness, so they are either par­al­lel or the same line. If also $c_1 = kc_2$, the first equa­tion is exactly $k$ times the sec­ond, so the two lines are iden­ti­cal. If $c_1 \neq kc_2$, then $a_2x + b_2y$ would have to equal two dif­fer­ent num­bers at once, which is impos­si­ble, so there is no com­mon point. If the first two ratios dif­fer, the lines have dif­fer­ent steep­ness and must cross exactly once.

Three pairs checked

Pair 1: $x - 2y = 0$ and $3x + 4y - 20 = 0$. Here $\displaystyle \dfrac{a_1}{a_2} = \dfrac{1}{3}$ and $\displaystyle \dfrac{b_1}{b_2} = \dfrac{-2}{4} = -\dfrac{1}{2}$. These dif­fer, so the lines inter­sect.

Pair 2: $2x + 3y - 9 = 0$ and $4x + 6y - 18 = 0$. Here $\displaystyle \dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$, $\displaystyle \dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}$ and $\displaystyle \dfrac{c_1}{c_2} = \dfrac{-9}{-18} = \dfrac{1}{2}$. All equal, so the lines coin­cide.

Pair 3: $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$. Here $\displaystyle \dfrac{a_1}{a_2} = \dfrac{1}{2}$ and $\displaystyle \dfrac{b_1}{b_2} = \dfrac{2}{4} = \dfrac{1}{2}$, but $\displaystyle \dfrac{c_1}{c_2} = \dfrac{-4}{-12} = \dfrac{1}{3}$. The first two agree and the third does not, so the lines are par­al­lel.

The graph­i­cal method

Method in steps

1. Write each equa­tion so that $y$ is easy to find, for exam­ple $\displaystyle y = \dfrac{c - ax}{b}$ for $ax + by = c$.
2. For each equa­tion, choose at least two val­ues of $x$ (three is safer) and find the match­ing $y$. Choose val­ues that give whole num­bers.
3. Plot the points on the same axes with a suit­able scale and draw a straight line through each set.
4. Read off the point where the lines meet. If they are par­al­lel, there is no solu­tion; if they coin­cide, there are infi­nitely many.
5. Check the point by sub­sti­tut­ing it into both orig­i­nal equa­tions.

Worked exam­ples

Exam­ple 1: Akhi­la's rides and games

Solve $x - 2y = 0$ and $3x + 4y = 20$ graph­i­cally.

Points on $x - 2y = 0$ (that is, $\displaystyle y = \dfrac{x}{2}$): $(0, 0)$ and $(2, 1)$.

Points on $3x + 4y = 20$: when $x = 0$, $y = 5$, giv­ing $(0, 5)$; when $x = 4$, $4y = 8$, so $y = 2$, giv­ing $(4, 2)$.

The point $(4, 2)$ sat­is­fies $\displaystyle y = \dfrac{x}{2}$ as well, so the two lines meet there.

Check: $4 - 2(2) = 0$ and $3(4) + 4(2) = 12 + 8 = 20$. In money terms, 4 rides cost ₹12 and 2 games cost ₹8, which is ₹20 in all.

Answer: Akhila took 4 rides on the Giant Wheel and played Hoopla 2 times.

Exam­ple 2: A con­sis­tent pair solved on the graph

Check whether $x + 3y = 6$ and $2x - 3y = 12$ are con­sis­tent. If so, solve them graph­i­cally.

Ratio test: $\displaystyle \dfrac{a_1}{a_2} = \dfrac{1}{2}$ and $\displaystyle \dfrac{b_1}{b_2} = \dfrac{3}{-3} = -1$. These dif­fer, so the pair is con­sis­tent with a unique solu­tion.

For $x + 3y = 6$: $x = 0$ gives $y = 2$, and $x = 6$ gives $y = 0$. Points $A(0, 2)$ and $B(6, 0)$.

For $2x - 3y = 12$: $x = 0$ gives $y = -4$, and $x = 3$ gives $-3y = 6$, so $y = -2$. Points $P(0, -4)$ and $Q(3, -2)$.

Draw­ing line $AB$ and line $PQ$, they meet at $B(6, 0)$.

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

Answer: the pair is con­sis­tent, and the solu­tion is $x = 6$, $y = 0$.

Exam­ple 3: Coin­ci­dent lines with­out draw­ing

Does the pair $5x - 8y + 1 = 0$ and $\displaystyle 3x - \dfrac{24}{5}y + \dfrac{3}{5} = 0$ have no solu­tion, one solu­tion or infi­nitely many?

Mul­ti­ply the sec­ond equa­tion by $\displaystyle \dfrac{5}{3}$:

$$\displaystyle \frac{5}{3}\left(3x - \frac{24}{5}y + \frac{3}{5}\right) = 5x - 8y + 1$$

This is exactly the first equa­tion, so the two equa­tions describe the same line. The ratio test agrees: $\displaystyle \dfrac{a_1}{a_2} = \dfrac{5}{3}$, $\displaystyle \dfrac{b_1}{b_2} = \dfrac{-8}{-24/5} = \dfrac{5}{3}$ and $\displaystyle \dfrac{c_1}{c_2} = \dfrac{1}{3/5} = \dfrac{5}{3}$.

Answer: the lines are coin­ci­dent, so the pair has infi­nitely many solu­tions.

Exam­ple 4: Cham­pa's shop­ping

Champa went to a "Sale" to buy some pants and skirts. When her friends asked how many of each she had bought, she said: "The num­ber of skirts is two less than twice the num­ber of pants pur­chased. Also, the num­ber of skirts is four less than four times the num­ber of pants pur­chased." How many of each did she buy?

Let $x$ be the num­ber of pants and $y$ the num­ber of skirts. Then

$$y = 2x - 2 \qquad \text{and} \qquad y = 4x - 4$$

Points on $y = 2x - 2$: $(0, -2)$ and $(2, 2)$. Points on $y = 4x - 4$: $(0, -4)$ and $(1, 0)$.

The point $(1, 0)$ is also on the first line, since $2(1) - 2 = 0$. So the lines meet at $(1, 0)$.

Check: twice 1 minus 2 is 0, and four times 1 minus 4 is 0. Both state­ments hold.

Answer: Champa bought 1 pair of pants and no skirts.

Exam­ple 5: Spot­ting a par­al­lel pair

Show that $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$ have no com­mon solu­tion.

Mul­ti­ply the first equa­tion by 2: $2x + 4y = 8$. The sec­ond equa­tion says $2x + 4y = 12$. The same expres­sion can­not equal both 8 and 12, so no point lies on both lines.

Answer: the lines are par­al­lel, and the pair is incon­sis­tent.

Where the graph­i­cal method falls short

The graph shows at a glance whether there is one solu­tion, none or infi­nitely many. Its weak­ness is pre­ci­sion. If the true solu­tion is some­thing like $(\sqrt{3}, 2\sqrt{7})$, $(-1.75, 3.3)$ or $\displaystyle \left(\dfrac{4}{13}, \dfrac{1}{19}\right)$, no pen­cil draw­ing can locate it exactly. That is why the chap­ter goes on to the alge­braic meth­ods, sub­sti­tu­tion and elim­i­na­tion, which give exact answers.

Com­mon mis­takes

• Plot­ting only two points and mis­read­ing one of them. Use a third point on each line as a check.
• Get­ting the signs of $c_1$ and $c_2$ wrong: $2x + 3y = 9$ becomes $2x + 3y - 9 = 0$, so $c_1 = -9$, not $9$.
• Call­ing a coin­ci­dent pair incon­sis­tent. A coin­ci­dent pair has infi­nitely many solu­tions, so it is con­sis­tent.
• Read­ing an inter­sec­tion point off the graph and not sub­sti­tut­ing it back into both equa­tions.
• Using unequal scales on the two axes with­out notic­ing, which makes the lines look wrongly placed.
• Stop­ping at "the lines meet at $(4, 2)$" in a word prob­lem instead of answer­ing the ques­tion in words.

Try these

1. Is the pair $3x + 2y = 7$ and $6x + 4y = 11$ con­sis­tent? Answer: No. The lines are par­al­lel, so there is no solu­tion.
2. Solve graph­i­cally: $x + y = 7$ and $x - y = 1$. Answer: $x = 4$, $y = 3$.
3. Solve graph­i­cally: $2x + y = 8$ and $x - y = 1$. Answer: $x = 3$, $y = 2$.
4. Clas­sify the pair $x - 3y = 4$ and $3x - 9y = 12$. Answer: coin­ci­dent lines, infi­nitely many solu­tions.
5. For what value of $k$ does the pair $kx + 2y = 5$ and $3x + y = 1$ fail to have a unique solu­tion? Answer: $k = 6$ (for every other value of $k$ the solu­tion is unique).
6. The sum of two num­bers is 12 and their dif­fer­ence is 4. Find the num­bers by draw­ing graphs. Answer: 8 and 4.

Key terms

Lin­ear equa­tion in two vari­ables

An equa­tion of the form $ax + by + c = 0$ with $a$ and $b$ not both zero; its graph is a straight line.

Pair of lin­ear equa­tions

Two lin­ear equa­tions in the same two vari­ables, con­sid­ered together.

Solu­tion of a pair

Val­ues of $x$ and $y$ that sat­isfy both equa­tions at once; on a graph, a com­mon point of the two lines.

Con­sis­tent pair

A pair with at least one solu­tion: inter­sect­ing or coin­ci­dent lines.

Incon­sis­tent pair

A pair with no solu­tion: par­al­lel lines.

Depen­dent pair

A pair whose two equa­tions describe the same line, giv­ing infi­nitely many solu­tions.

Coin­ci­dent lines

Two lines that lie exactly on top of each other.

Ratio test

Com­par­ing $\displaystyle \dfrac{a_1}{a_2}$, $\displaystyle \dfrac{b_1}{b_2}$ and $\displaystyle \dfrac{c_1}{c_2}$ to decide which of the three cases a pair belongs to.

Com­mon ques­tions

How many points do I need to draw a line?

Two points fix a line, but plot­ting a third is a good habit. If the three points do not lie in a straight line, there is an arith­metic slip some­where.

Is a depen­dent pair con­sis­tent?

Yes. Con­sis­tent means the pair has at least one solu­tion, and a depen­dent pair has infi­nitely many.

What if one of $a_2$, $b_2$ or $c_2$ is zero?

Then the ratio can­not be formed directly. Com­pare the equa­tions instead by check­ing whether one is a mul­ti­ple of the other, or draw the lines. For exam­ple, $y = 2$ and $y = 5$ are both hor­i­zon­tal, so they are par­al­lel.

Can the answer to a word prob­lem be a frac­tion?

It depends on the con­text. Rupees per item can be a frac­tion, but a num­ber of peo­ple or rides must be a whole num­ber. If you get a frac­tion where only whole num­bers make sense, recheck the equa­tions.

Why learn the graph­i­cal method if alge­bra is more exact?

The graph explains why a pair can have one, none or infi­nitely many solu­tions, and it gives a quick visual check on an alge­braic answer.

Ref­er­ences

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Stew­art, J., Redlin, L. and Wat­son, S. Pre­cal­cu­lus: Math­e­mat­ics for Cal­cu­lus. Cen­gage Learn­ing.
3. Gelfand, I. M. and Shen, A. Alge­bra. Birkhäuser.