Linear Equations in Two Variables

Recognise an equation of the form $ax + by = c$ , check whether a pair of values $(x, y)$ is a solution, find solutions and tables of values, and write such an equation from words.

Before you start

To solve a linear equation in one unknown, do the same to both sides: $3y + 6 = 12$ gives $3y = 6$ , so $y = 2$ .
To substitute a negative number, use brackets: when $x = -3$ , $2x = 2(-3) = -6$ .
To make $y$ the subject of a formula, rearrange it so that $y$ stands alone on one side.

What Is a Linear Equation in Two Variables?

In Secondary 1 an equation such as $3x + 2 = 11$ had one unknown, and one value of $x$ made it true. Many situations need two unknowns. If a pen costs $ $x$ and a file costs $ $y$ , and 3 pens and 2 files cost $ $12$ , then $3x + 2y = 12$ . This is an equation in two variables, $x$ and $y$ .

Linear equation in two variables

A linear equation in two variables $x$ and $y$ is an equation that can be written in the form

$ax + by = c,$

where $a$ , $b$ and $c$ are numbers and $a$ and $b$ are not both $0$ .

Each variable appears only to the power $1$ . There is no $x^{2}$ , no $xy$ (the variables are not multiplied together) and no variable in a denominator, such as $\frac{1}{x}$ .

Example 1: Is it a linear equation in two variables?

Which of these are linear equations in two variables? Give a reason for each. (a) $3x - 2y = 6$ (b) $y = 4x - 1$ (c) $xy = 6$ (d) $x^{2} + y = 4$ (e) $\frac{1}{x} + y = 2$ (f) $2(x + y) = 3x - 4$ (g) $\frac{x}{2} + \frac{y}{3} = 1$

Solution

(a) Yes. It is already in the form $ax + by = c$ , with $a = 3$ , $b = -2$ and $c = 6$ .

(b) Yes. Subtract $4x$ from both sides: $-4x + y = -1$ . This is the form $ax + by = c$ with $a = -4$ , $b = 1$ , $c = -1$ .

(d) No. $x$ is squared.

(e) No. $x$ is in a denominator: $\frac{1}{x}$ is not a number times $x$ .

(f) Yes. Expand the bracket: $2x + 2y = 3x - 4$ . Subtract $3x$ from both sides: $-x + 2y = -4$ . This is the form $ax + by = c$ with $a = -1$ , $b = 2$ and $c = -4$ .

(g) Yes. $\frac{x}{2}$ is $\frac{1}{2}x$ and $\frac{y}{3}$ is $\frac{1}{3}y$ , so $a = \frac{1}{2}$ , $b = \frac{1}{3}$ and $c = 1$ . Fractions are allowed as the numbers $a$ , $b$ and $c$ .

(a), (b), (f) and (g) are linear equations in two variables; (c), (d) and (e) are not.

Solutions Are Ordered Pairs

A solution of an equation in two variables

A solution of an equation in $x$ and $y$ is a pair of values, one for $x$ and one for $y$ , that makes the equation true.

It is written as an ordered pair $(x, y)$ , with the value of $x$ first. $(3, 2)$ means $x = 3$ and $y = 2$ .

To check a pair, substitute both values and see whether the two sides are equal.

Example 2: Checking pairs

Which of the pairs $(3, 2)$ , $(6, 0)$ , $(2, 3)$ and $(-3, 6)$ are solutions of $2x + 3y = 12$ ?

Solution

Substitute each pair into the left-hand side, $2x + 3y$ , and compare with $12$ .

$(3, 2)$ : $2(3) + 3(2) = 6 + 6 = 12\text{.}$ Equal to $12$ , so it is a solution.

$(6, 0)$ : $2(6) + 3(0) = 12 + 0 = 12\text{.}$ A solution.

$(2, 3)$ : $2(2) + 3(3) = 4 + 9 = 13\text{.}$ Not equal to $12$ , so it is not a solution.

$(-3, 6)$ : $2(-3) + 3(6) = -6 + 18 = 12\text{.}$ A solution.

Look at $(3, 2)$ and $(2, 3)$ : the same two numbers, but only one is a solution. The order matters, which is why the pair is called ordered.

$(3, 2)$ , $(6, 0)$ and $(-3, 6)$ are solutions of $2x + 3y = 12$ ; $(2, 3)$ is not.

Finding a Solution from One Value

If you know one of the two values, substitute it. What is left is an equation in one unknown, which you solve as in Secondary 1. The answer may be negative or a fraction.

Example 3: Given x, find y; given y, find x

For the equation $2x + 3y = 12$ , find (a) $y$ when $x = 9$ , (b) $y$ when $x = 1$ , (c) $x$ when $y = 5$ , (d) $x$ when $y = 1$ .

Solution

(a) $2(9) + 3y = 12$ , so $18 + 3y = 12$ . Subtract $18$ : $3y = -6$ . Divide by $3$ : $y = -2$ .

(b) $2(1) + 3y = 12$ , so $2 + 3y = 12$ . Subtract $2$ : $3y = 10$ . Divide by $3$ : $y = \frac{10}{3} = 3\frac{1}{3}$ .

(d) $2x + 3(1) = 12$ , so $2x + 3 = 12$ . Subtract $3$ : $2x = 9$ . Divide by $2$ : $x = 4.5$ .

Check (a): $2(9) + 3(-2) = 18 - 6 = 12\text{.}$ Check (c): $2(-1.5) + 3(5) = -3 + 15 = 12\text{.}$

(a) $y = -2$ , so $(9, -2)$ is a solution. (b) $y = 3\frac{1}{3}$ , so $(1, 3\frac{1}{3})$ . (c) $x = -1.5$ , so $(-1.5, 5)$ . (d) $x = 4.5$ , so $(4.5, 1)$ .

Infinitely Many Solutions

Whatever value you choose for $x$ , you can find the value of $y$ that goes with it, as in Example 3. There is no end to the values you can choose, so one linear equation in two variables has infinitely many solutions.

To find solutions quickly, choose values that make the arithmetic easy, such as $0$ , or values that make the division come out whole.

Example 4: Listing solutions

Find five solutions of $x - 2y = 4$ . Include one in which $x$ is negative and one in which $y$ is a fraction.

Solution

Here it is easier to choose $y$ and find $x$ : add $2y$ to both sides to get $x = 4 + 2y$ .

$y = 0$ : $x = 4 + 0 = 4$ , giving $(4, 0)$ .

$y = 1$ : $x = 4 + 2 = 6$ , giving $(6, 1)$ .

$y = -1$ : $x = 4 + 2(-1) = 2$ , giving $(2, -1)$ .

$y = -3$ : $x = 4 + 2(-3) = -2\text{,}$ giving $(-2, -3)$ , with $x$ negative.

For a fraction, choose $x = 5$ : $5 - 2y = 4$ , so $-2y = -1$ and $y = \frac{1}{2}$ , giving $(5, \frac{1}{2})$ .

Check $(-2, -3)$ : $-2 - 2(-3) = -2 + 6 = 4\text{.}$ Check $(5, \frac{1}{2})$ : $5 - 2(\frac{1}{2}) = 5 - 1 = 4\text{.}$

For example $(4, 0)$ , $(6, 1)$ , $(2, -1)$ , $(-2, -3)$ and $(5, \frac{1}{2})$ . There are infinitely many others.

The drawing shows seven solutions of $2x + 3y = 12$ plotted as points: six found in Examples 2 and 3, and $(0, 4)$ , since $2(0) + 3(4) = 12$ . They lie on one straight line. Every solution of the equation is a point of that line: this is the graph of the equation, which the next lesson draws.

Making y the Subject and Tables of Values

To find $y$ for many values of $x$ , it saves work to make $y$ the subject first. Then each value of $x$ is substituted into one expression.

Making y the subject of ax + by = c

Subtract $ax$ from both sides: $by = c - ax$ .

Divide both terms by $b$ (this needs $b \ne 0$ ): $y = \frac{c - ax}{b}.$

Take care with signs when $b$ is negative.

Example 5: A table of values for 5x + 2y = 10

Make $y$ the subject of $5x + 2y = 10$ , and use it to complete a table of values for $x = -2, -1, 0, 1, 2$ .

Solution

Subtract $5x$ from both sides: $2y = 10 - 5x$ . Divide by $2$ : $y = \frac{10 - 5x}{2}$ .

$x = -2$ : $y = \frac{10 - 5(-2)}{2} = \frac{10 + 10}{2} = 10$

$x = -1$ : $y = \frac{10 + 5}{2} = \frac{15}{2} = 7.5$

$x = 0$ : $y = \frac{10}{2} = 5$

$x = 1$ : $y = \frac{10 - 5}{2} = 2.5$

$x = 2$ : $y = \frac{10 - 10}{2} = 0$

$x$	$-2$	$-1$	$0$	$1$	$2$
$y$	$10$	$7.5$	$5$	$2.5$	$0$

Check $x = -1$ in the original equation: $5(-1) + 2(7.5) = -5 + 15 = 10\text{.}$

$y = \frac{10 - 5x}{2}$ , and the values of $y$ are $10$ , $7.5$ , $5$ , $2.5$ and $0$ .

Example 6: When b is negative

Make $y$ the subject of $2x - 3y = 6$ , and find $y$ for $x = -3, 0, 3, 6$ .

Solution

Subtract $2x$ from both sides: $-3y = 6 - 2x$ .

Multiply both sides by $-1$ to make the $y$ term positive: $3y = 2x - 6$ . Divide by $3$ : $y = \frac{2x - 6}{3}$ .

$x = -3$ : $y = \frac{2(-3) - 6}{3} = \frac{-12}{3} = -4$

$x = 0$ : $y = \frac{-6}{3} = -2$

$x = 3$ : $y = \frac{6 - 6}{3} = 0$

$x = 6$ : $y = \frac{12 - 6}{3} = 2$

$x$	$-3$	$0$	$3$	$6$
$y$	$-4$	$-2$	$0$	$2$

Check $x = -3$ : $2(-3) - 3(-4) = -6 + 12 = 6\text{.}$

$y = \frac{2x - 6}{3}$ , and the values of $y$ are $-4$ , $-2$ , $0$ and $2$ .

Finding an Unknown in a Solution

If you are told that a pair is a solution, its values make the equation true. Substitute them, and solve for the unknown.

Example 7: Finding k

(a) $(k, 2)$ is a solution of $3x - 4y = 7$ . Find $k$ . (b) $(3, k)$ is a solution of $5x + 2y = 9$ . Find $k$ .

Solution

(a) Here $x = k$ and $y = 2$ : $3k - 4(2) = 7$ , so $3k - 8 = 7$ . Add $8$ : $3k = 15$ . Divide by $3$ : $k = 5$ .

Check: $3(5) - 4(2) = 15 - 8 = 7\text{.}$

(b) Here $x = 3$ and $y = k$ : $5(3) + 2k = 9$ , so $15 + 2k = 9$ . Subtract $15$ : $2k = -6$ . Divide by $2$ : $k = -3$ .

Check: $5(3) + 2(-3) = 15 - 6 = 9\text{.}$

(a) $k = 5$ (b) $k = -3$

Equations with the Same Solutions

Equivalent equations

Multiplying or dividing every term on both sides of an equation by the same number (not $0$ ), or rearranging its terms, gives an equation with exactly the same solutions. Such equations are called equivalent.

$2x + 4y = 10$ and $x + 2y = 5$ are equivalent: divide every term of the first by $2$ to get the second.

Example 8: Which equations are equivalent?

Which of these equations have exactly the same solutions as $3x - y = 4$ ? (a) $6x - 2y = 8$ (b) $y = 3x - 4$ (c) $3x - y = 8$ (d) $9x - 3y = 4$

Solution

(a) Every term of $3x - y = 4$ multiplied by $2$ gives $6x - 2y = 8$ . Equivalent.

(b) Add $y$ to both sides of $3x - y = 4$ , then subtract $4$ : $3x - 4 = y$ , that is $y = 3x - 4$ . Equivalent.

(c) Only the right-hand side has changed. $(2, 2)$ is a solution of $3x - y = 4$ , since $6 - 2 = 4$ , but not of $3x - y = 8$ . Not equivalent.

(d) The left-hand side was multiplied by $3$ but the right-hand side was not. $(2, 2)$ gives $18 - 6 = 12$ , not $4$ . Not equivalent.

(a) and (b) are equivalent to $3x - y = 4$ ; (c) and (d) are not.

Writing an Equation from Words

First say what each letter stands for, with its unit. Then turn the sentence into an equation.

Example 9: From a sentence to an equation

Write an equation in $x$ and $y$ for each sentence. (a) The sum of two numbers is $10$ . (b) The difference between two numbers is $4$ , the first being the larger. (c) 3 pens and 2 files cost $ $12$ . (d) The chickens and goats on a farm have $50$ legs altogether.

Solution

(a) Let the numbers be $x$ and $y$ : $x + y = 10$ .

(b) Let the larger number be $x$ and the smaller $y$ : $x - y = 4$ .

(d) Let there be $x$ chickens and $y$ goats. A chicken has 2 legs and a goat 4: $2x + 4y = 50$ .

(a) $x + y = 10$ (b) $x - y = 4$ (c) $3x + 2y = 12$ (d) $2x + 4y = 50$

In a real setting, not every solution makes sense. You cannot have $2.5$ goats or $-1$ notes. When the variables count things, only whole numbers that are not negative will do, and there are usually only a few such solutions.

Example 10: Whole-number solutions

Mrs Lim pays exactly $ $30$ using only $ $2$ notes and $ $5$ notes, with at least one note of each kind. Write an equation and list every way she can do it.

Solution

Let $x$ be the number of $ $2$ notes and $y$ the number of $ $5$ notes: $2x + 5y = 30$ .

Make $x$ the subject: $x = \frac{30 - 5y}{2}$ . For $x$ to be a whole number, $30 - 5y$ must be even, so $y$ must be even.

Try the even values of $y$ from $2$ up. Stop when $x$ would be $0$ or less.

$y$	$2$	$4$	$6$
$x = \frac{30 - 5y}{2}$	$10$	$5$	$0$

$y = 6$ gives $x = 0$ , which has no $ $2$ notes, so it is not allowed.

Check: $2(10) + 5(2) = 20 + 10 = 30$ and $2(5) + 5(4) = 10 + 20 = 30\text{.}$

There are two ways: ten $ $2$ notes and two $ $5$ notes, or five $ $2$ notes and four $ $5$ notes.

Common Mistakes

Swapping the order. $(2, 3)$ means $x = 2$ and $y = 3$ . In Example 2, $(3, 2)$ is a solution of $2x + 3y = 12$ but $(2, 3)$ is not.
Losing a minus sign. Substitute a negative number in brackets: $2x - 3y$ at $y = -4$ is $2x - 3(-4) = 2x + 12\text{,}$ not $2x - 12$ .
Dividing only one term. From $2y = 10 - 5x$ , $y = \frac{10 - 5x}{2}$ , not $10 - \frac{5x}{2}$ . Both terms are divided by $2$ .
Changing one side only. $2x + 4y = 5$ is not equivalent to $x + 2y = 5$ : every term on both sides must be divided by $2$ .
Calling $xy = 6$ linear. A term with $x$ times $y$ , or with $x^{2}$ , or with $x$ in a denominator, makes the equation not linear.

Practice Questions

Work each question on paper before you open its solution. The questions get harder as you go.

Question 1

Which of these are linear equations in two variables? (a) $5x - y = 2$ (b) $x^{2} - 3y = 1$ (c) $\frac{4}{y} + x = 3$ (d) $y = 7 - 2x$ (e) $3xy = 12$ (f) $x(y + 2) = 5$

ShowHide solution

(a) Yes: the form $ax + by = c$ with $a = 5$ , $b = -1$ , $c = 2$ .

(b) No: $x$ is squared.

(d) Yes: add $2x$ to both sides to get $2x + y = 7$ .

(e) No: $x$ and $y$ are multiplied together.

(f) No: expanding gives $xy + 2x = 5$ , and $x$ and $y$ are multiplied together.

(a) and (d) only.

Question 2

Which of the pairs $(2, 1)$ , $(1, 2)$ , $(4, 7)$ and $(0, -3)$ are solutions of $3x - y = 5$ ?

ShowHide solution

$(2, 1)$ : $3(2) - 1 = 6 - 1 = 5$ . Yes.

$(1, 2)$ : $3(1) - 2 = 1$ . No.

$(4, 7)$ : $3(4) - 7 = 12 - 7 = 5\text{.}$ Yes.

$(0, -3)$ : $3(0) - (-3) = 0 + 3 = 3\text{.}$ No.

$(2, 1)$ and $(4, 7)$ are solutions; $(1, 2)$ and $(0, -3)$ are not.

Question 3

For the equation $4x - 5y = 20$ , find (a) $y$ when $x = 0$ , (b) $x$ when $y = 2$ , (c) $y$ when $x = -5$ , (d) $y$ when $x = 3$ .

ShowHide solution

(a) $4(0) - 5y = 20$ , so $-5y = 20$ and $y = -4$ .

(b) $4x - 5(2) = 20$ , so $4x - 10 = 20$ , $4x = 30$ and $x = 7.5$ .

(d) $4(3) - 5y = 20$ , so $12 - 5y = 20$ . Subtract $12$ : $-5y = 8$ , so $y = -\frac{8}{5} = -1.6$ .

Check (c): $4(-5) - 5(-8) = -20 + 40 = 20\text{.}$

(a) $y = -4$ (b) $x = 7.5$ (c) $y = -8$ (d) $y = -1.6$

Question 4

Make $y$ the subject of $4x + 3y = 12$ . Use it to complete the table of values for $x = -3, 0, 3, 6$ .

ShowHide solution

Subtract $4x$ : $3y = 12 - 4x$ . Divide by $3$ : $y = \frac{12 - 4x}{3}$ .

$x = -3$ : $y = \frac{12 + 12}{3} = 8$ . $x = 0$ : $y = \frac{12}{3} = 4$ . $x = 3$ : $y = \frac{12 - 12}{3} = 0$ . $x = 6$ : $y = \frac{12 - 24}{3} = -4\text{.}$

$x$	$-3$	$0$	$3$	$6$
$y$	$8$	$4$	$0$	$-4$

Check $x = 6$ : $4(6) + 3(-4) = 24 - 12 = 12\text{.}$

$y = \frac{12 - 4x}{3}$ ; the values of $y$ are $8$ , $4$ , $0$ and $-4$ .

Question 5

Make $y$ the subject of $6x - 2y = 9$ . Use it to find $y$ for $x = -1, 0, 1, 2$ .

ShowHide solution

Subtract $6x$ : $-2y = 9 - 6x$ . Multiply both sides by $-1$ : $2y = 6x - 9$ . Divide by $2$ : $y = \frac{6x - 9}{2}$ .

$x = -1$ : $y = \frac{-6 - 9}{2} = -7.5\text{.}$ $x = 0$ : $y = -4.5$ . $x = 1$ : $y = \frac{-3}{2} = -1.5$ . $x = 2$ : $y = \frac{3}{2} = 1.5$ .

Check $x = -1$ : $6(-1) - 2(-7.5) = -6 + 15 = 9\text{.}$

$y = \frac{6x - 9}{2}$ ; the values of $y$ are $-7.5$ , $-4.5$ , $-1.5$ and $1.5$ .

Question 6

Complete this table for the equation $2x + 3y = 6$ . Some of the missing values are values of $x$ .

$x$	$-3$	$0$	?	?	$1.5$
$y$	?	?	$0$	$-2$	?

ShowHide solution

$x = -3$ : $-6 + 3y = 6$ , so $3y = 12$ and $y = 4$ .

$x = 0$ : $3y = 6$ , so $y = 2$ .

$y = 0$ : $2x = 6$ , so $x = 3$ .

$y = -2$ : $2x - 6 = 6$ , so $2x = 12$ and $x = 6$ .

$x = 1.5$ : $3 + 3y = 6$ , so $3y = 3$ and $y = 1$ .

The pairs are $(-3, 4)$ , $(0, 2)$ , $(3, 0)$ , $(6, -2)$ and $(1.5, 1)$ .

Question 7

Write down four solutions of $x + 3y = 9$ : one with a negative value of $x$ , and one in which $y$ is not a whole number.

ShowHide solution

Make $x$ the subject: $x = 9 - 3y$ . Choose values of $y$ .

$y = 0$ : $x = 9$ , giving $(9, 0)$ . $y = 3$ : $x = 0$ , giving $(0, 3)$ . $y = 4$ : $x = 9 - 12 = -3$ , giving $(-3, 4)$ .

For $y$ not whole, choose $x = 1$ : $1 + 3y = 9$ , so $3y = 8$ and $y = \frac{8}{3} = 2\frac{2}{3}$ , giving $(1, 2\frac{2}{3})$ .

Check $(-3, 4)$ : $-3 + 12 = 9$ . Check $(1, 2\frac{2}{3})$ : $1 + 3 \times \frac{8}{3} = 1 + 8 = 9\text{.}$

For example $(9, 0)$ , $(0, 3)$ , $(-3, 4)$ and $(1, 2\frac{2}{3})$ . Any pair that makes $x + 3y = 9$ true is right.

Question 8

(a) $(k, -1)$ is a solution of $2x + 5y = 7$ . Find $k$ . (b) $(-2, k)$ is a solution of $3x - 2y = 4$ . Find $k$ . (c) $(p, p)$ is a solution of $3x + 2y = 15$ . Find $p$ .

ShowHide solution

(a) $2k + 5(-1) = 7$ , so $2k - 5 = 7$ , $2k = 12$ and $k = 6$ . Check: $12 - 5 = 7$ .

(b) $3(-2) - 2k = 4$ , so $-6 - 2k = 4$ . Add $6$ : $-2k = 10$ , so $k = -5$ . Check: $-6 - 2(-5) = -6 + 10 = 4\text{.}$

(a) $k = 6$ (b) $k = -5$ (c) $p = 3$

Question 9

Which of these equations have exactly the same solutions as $x - 2y = 3$ ? (a) $2x - 4y = 6$ (b) $3x - 6y = 3$ (c) $x = 2y + 3$ (d) $-x + 2y = -3$ (e) $2x - 2y = 6$

ShowHide solution

(a) Every term multiplied by $2$ . Equivalent.

(b) The left-hand side is multiplied by $3$ but $3 \times 3 = 9$ , not $3$ . Not equivalent: $(3, 0)$ solves $x - 2y = 3$ but gives $9 - 0 = 9$ here.

(d) Every term multiplied by $-1$ . Equivalent.

(e) $x$ was multiplied by $2$ but $-2y$ was not. Not equivalent: $(3, 0)$ works, but $(5, 1)$ solves $x - 2y = 3$ and gives $10 - 2 = 8$ here.

(a), (c) and (d).

Question 10

Write an equation in $x$ and $y$ for each. (a) A number $x$ is $5$ less than three times a number $y$ . (b) At a cinema, $x$ adult tickets at $ $12$ each and $y$ child tickets at $ $8$ each cost $ $96$ altogether. (c) A rectangle $x$ cm long and $y$ cm wide has a perimeter of $26$ cm.

ShowHide solution

(a) Three times $y$ is $3y$ ; $5$ less is $3y - 5$ : $x = 3y - 5$ , or $x - 3y = -5$ .

(b) The adult tickets cost $ $12x$ and the child tickets $ $8y$ : $12x + 8y = 96$ . Dividing every term by $4$ gives the equivalent $3x + 2y = 24$ .

(a) $x = 3y - 5$ (b) $12x + 8y = 96$ (c) $2x + 2y = 26$

Question 11

Kumar spends exactly $ $40$ on pens at $ $3$ each and notebooks at $ $4$ each, and buys at least one of each. Write an equation and find every possible number of pens and notebooks.

ShowHide solution

Let $x$ be the number of pens and $y$ the number of notebooks: $3x + 4y = 40$ , so $x = \frac{40 - 4y}{3}$ .

Try $y = 1, 2, 3, \ldots$ and keep only whole values of $x$ that are at least $1$ :

$y$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$40 - 4y$	$36$	$32$	$28$	$24$	$20$	$16$	$12$	$8$	$4$	$0$
$x$	$12$	—	—	$8$	—	—	$4$	—	—	$0$

$y = 10$ gives $x = 0$ : no pens, so it is not allowed. A dash means $40 - 4y$ is not a multiple of $3$ .

Check: $3(12) + 4(1) = 40$ , $3(8) + 4(4) = 24 + 16 = 40\text{,}$ $3(4) + 4(7) = 12 + 28 = 40\text{.}$

Three ways: 12 pens and 1 notebook, 8 pens and 4 notebooks, or 4 pens and 7 notebooks.

Question 12

Aisha has only 20-cent and 50-cent coins in her purse, worth $ $2$ altogether. She has at least one coin of each kind. How many of each could she have?

ShowHide solution

Work in cents: $ $2$ is $200$ cents. Let there be $x$ 20-cent coins and $y$ 50-cent coins: $20x + 50y = 200$ .

Divide every term by $10$ : $2x + 5y = 20$ , so $x = \frac{20 - 5y}{2}$ . For a whole $x$ , $y$ must be even.

$y = 2$ : $x = \frac{20 - 10}{2} = 5$ . $y = 4$ : $x = 0$ , not allowed (no 20-cent coins).

Check: $5$ coins of 20 cents make $100$ cents and $2$ coins of 50 cents make $100$ cents; together $200$ cents, or $ $2$ .

Only one way: five 20-cent coins and two 50-cent coins.

Key Points

A linear equation in two variables can be written as $ax + by = c$ , with $a$ and $b$ not both $0$ ; no $x^{2}$ , no $xy$ , no variable in a denominator.
A solution is an ordered pair $(x, y)$ , $x$ first, that makes the equation true. Check a pair by substituting both values.
Given one value, substitute it and solve for the other. The answer may be negative or a fraction.
One linear equation in two variables has infinitely many solutions.
To make a table of values, first make $y$ the subject: $y = \frac{c - ax}{b}$ (when $b \ne 0$ ).
If a pair with an unknown $k$ is a solution, substitute it and solve for $k$ .
Multiplying or dividing every term by the same non-zero number gives an equivalent equation, with the same solutions.
In a setting where the variables count things, only whole numbers that are not negative make sense.