Pairs of equations and their solutions

One equa­tion in two vari­ables has infi­nitely many solu­tions. To find a sin­gle answer you need two equa­tions that are true at the same time. This les­son explains what such a pair is, what its solu­tion means on a graph, and how to tell, just from the num­bers in front of $x$ and $y$, whether the pair has one solu­tion, none, or infi­nitely many, before you start solv­ing.

A pair of lin­ear equa­tions#

Two lin­ear equa­tions in the same two vari­ables, con­sid­ered together, form a pair of lin­ear equa­tions in two vari­ables. In gen­eral form a pair is writ­ten

$$\begin{aligned} a_1x + b_1y + c_1 &= 0 \\ a_2x + b_2y + c_2 &= 0 \end{aligned}$$

The small num­bers $1$ and $2$ writ­ten low down (sub­scripts) are labels, not pow­ers: $a_1$ is "the $a$ of the first equa­tion" and $a_2$ is "the $a$ of the sec­ond".

A solu­tion of the pair is an ordered pair $(x, y)$ that sat­is­fies both equa­tions. On a graph each equa­tion is a straight line, so a solu­tion of the pair is a point that lies on both lines: a point where the lines meet.

Exam­ple 1: solv­ing by draw­ing#

Find the solu­tion of the pair $x + y = 5$ and $x - y = 1$ by mak­ing tables.

For $x + y = 5$, that is $y = 5 - x$:

For $x - y = 1$, that is $y = x - 1$:

The point $(3, 2)$ appears in both tables. Drawn on a graph, the two lines cross at $(3, 2)$, so the solu­tion is $x = 3$, $y = 2$. Check: $3 + 2 = 5$ and $3 - 2 = 1$.

Draw­ing works well when the answer is a pair of whole num­bers. When the answer is some­thing like $\displaystyle \left(\frac{29}{13}, -\frac{11}{13}\right)$ no draw­ing is accu­rate enough, which is why the next three lessons teach meth­ods that use alge­bra instead.

Three ways two lines can sit#

Two straight lines on the same graph can be arranged in only three ways.

1. Inter­sect­ing lines. The lines cross at exactly one point. The pair has a unique solu­tion (one com­mon point). This hap­pens when

$$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$

2. Par­al­lel lines. The lines run side by side and never meet. The pair has no solu­tion. This hap­pens when

$$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$

3. Coin­ci­dent lines. The two equa­tions draw the same line, one lying exactly on top of the other. Every point of the line is on both, so the pair has infi­nitely many solu­tions. This hap­pens when

$$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

Why do the ratios decide this? The num­bers $a$ and $b$ fix how steeply a line slopes. If $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2}$ the two lines have the same slope, so they are either par­al­lel or the same line, and $c$ decides which. If the ratios dif­fer, the slopes dif­fer, and the lines must cross some­where.

A pair with at least one solu­tion (cases 1 and 3) is called con­sis­tent. A pair with no solu­tion (case 2) is called incon­sis­tent.

Exam­ple 2#

How many solu­tions does the pair $3x + 2y = 5$, $2x - 3y = 7$ have?

First write both in gen­eral form: $3x + 2y - 5 = 0$ and $2x - 3y - 7 = 0$. So $a_1 = 3$, $b_1 = 2$, $a_2 = 2$, $b_2 = -3$.

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

These are not equal, so the lines inter­sect and the pair has a unique solu­tion. You will find it, $\displaystyle \left(\frac{29}{13}, -\frac{11}{13}\right)$, in the next les­son.

Exam­ple 3#

How many solu­tions does the pair $2x + 3y - 8 = 0$, $4x + 6y - 16 = 0$ have?

$$\displaystyle \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$$

All three ratios are equal, so the lines coin­cide and there are infi­nitely many solu­tions. Indeed, the sec­ond equa­tion is just the first mul­ti­plied by $2$.

Exam­ple 4#

How many solu­tions does the pair $x - 2y + 3 = 0$, $3x - 6y + 5 = 0$ have?

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

The first two ratios are equal but the third is dif­fer­ent, so the lines are par­al­lel and the pair has no solu­tion.

Exam­ple 5#

For what value of $k$ does the pair $2x + ky = 3$, $4x + 6y = 5$ have no solu­tion?

For no solu­tion we need $\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2}$:

$$\displaystyle \begin{aligned} \frac{2}{4} &= \frac{k}{6} \\ 4k &= 12 \\ k &= 3 \end{aligned}$$

We must also check that $\displaystyle \frac{c_1}{c_2}$ is dif­fer­ent. In gen­eral form the con­stants are $-3$ and $-5$, and $\displaystyle \frac{-3}{-5} = \frac{3}{5}$, which is not $\displaystyle \frac{1}{2}$. So $k = 3$ gives no solu­tion.

Exam­ple 6#

For what val­ues of $k$ does the pair $kx + 2y = 5$, $3x + y = 1$ have a unique solu­tion?

We need $\displaystyle \frac{k}{3} \neq \frac{2}{1}$, so $k \neq 6$. Every value of $k$ except $6$ gives a unique solu­tion.

Prac­tice#

For ques­tions 1 to 5, say whether the pair has a unique solu­tion, no solu­tion, or infi­nitely many solu­tions.

1. $x + 2y - 4 = 0$ and $3x - y + 2 = 0$
2. $2x - 3y + 6 = 0$ and $4x - 6y + 12 = 0$
3. $5x + 2y = 7$ and $10x + 4y = 9$
4. $3x - y = 2$ and $6x - 2y = 4$
5. $4x + 5y = 20$ and $2x - y = 3$
6. For what val­ues of $k$ does $3x + ky = 4$, $6x + 8y = 1$ have a unique solu­tion?
7. Find $k$ so that $kx + 3y = 7$, $8x + 6y = 5$ has no solu­tion.
8. Find $k$ so that $2x + 3y = 7$, $4x + ky = 14$ has infi­nitely many solu­tions.
9. Use tables to find where the lines $x + y = 6$ and $x - y = 2$ meet.
10. Show that $(2, 1)$ is the solu­tion of the pair $x + 3y = 5$, $2x - y = 3$.

Answers#