Cross-multiplication and reducible equations

A formula for every pair

Solve the general pair

by elimination once, and the answer is a formula that works for every pair. Multiply the first equation by $b_2$ and the second by $b_1$:

Subtract the second from the first. The terms in $y$ cancel, and $(a_1b_2 - a_2b_1)x = b_1c_2 - b_2c_1$. The same steps for $y$ give the other result. Written together:

This is the cross-multiplication method. It works whenever the number under the $1$ is not zero.

How to use it

1. Write both equations as $ax + by + c = 0$, with everything on the left.
2. Read the six numbers $a_1, b_1, c_1$ and $a_2, b_2, c_2$, each with its sign.
3. Work out three numbers: $X = b_1c_2 - b_2c_1$, $Y = c_1a_2 - c_2a_1$ and $D = a_1b_2 - a_2b_1$.
4. Divide: $x = \dfrac{X}{D}$ and $y = \dfrac{Y}{D}$. If $D = 0$, the formula cannot be used.
5. Check the answer in both equations.

Each of the three numbers is one diagonal of a small square of coefficients, minus the other diagonal. The squares are $b$ and $c$ for $x$, then $c$ and $a$ for $y$, then $a$ and $b$ for the number under $1$.

A worked example

Solve $5x + 2y = 26$ and $3x - y = 9$.

1. Write them as $ax + by + c = 0$: $5x + 2y - 26 = 0$ and $3x - y - 9 = 0$.
2. Read the numbers. $a_1 = 5$, $b_1 = 2$, $c_1 = -26$, and $a_2 = 3$, $b_2 = -1$, $c_2 = -9$.
3. Work out the three numbers. $X = 2(-9) - (-1)(-26) = -18 - 26 = -44$. $Y = (-26)(3) - (-9)(5) = -78 + 45 = -33$. $D = 5(-1) - 3(2) = -11$.
4. Divide. $x = \dfrac{-44}{-11} = 4$ and $y = \dfrac{-33}{-11} = 3$.
5. Check. $5 \times 4 + 2 \times 3 = 26$ and $3 \times 4 - 3 = 9$.

The signs are the usual trouble. The constant $c$ is the negative of the number on the right of the equals sign.

Equations that become linear

Look at $\dfrac{2}{x} + \dfrac{3}{y} = 13$. It is not linear in $x$ and $y$, because they are in denominators. But it is linear in $\dfrac{1}{x}$ and $\dfrac{1}{y}$. So let $u = \dfrac{1}{x}$ and $v = \dfrac{1}{y}$.

Solve $\dfrac{2}{x} + \dfrac{3}{y} = 13$ and $\dfrac{5}{x} - \dfrac{4}{y} = -2$.

1. Change letters. The pair becomes $2u + 3v = 13$ and $5u - 4v = -2$.
2. Solve the new pair. Multiply the first by $4$ and the second by $3$: $8u + 12v = 52$ and $15u - 12v = -6$. Add them: $23u = 46$, so $u = 2$. Then $2 \times 2 + 3v = 13$ gives $v = 3$.
3. Go back to $x$ and $y$. $x = \dfrac{1}{u} = \dfrac{1}{2}$ and $y = \dfrac{1}{v} = \dfrac{1}{3}$.
4. Check. $\dfrac{2}{1/2} + \dfrac{3}{1/3} = 4 + 9 = 13$ and $\dfrac{5}{1/2} - \dfrac{4}{1/3} = 10 - 12 = -2$.

Here $x$ and $y$ cannot be $0$, because we divide by them.

A sum and a difference in the denominators

Solve $\dfrac{10}{x + y} + \dfrac{2}{x - y} = 4$ and $\dfrac{15}{x + y} - \dfrac{5}{x - y} = -2$.

Let $u = \dfrac{1}{x + y}$ and $v = \dfrac{1}{x - y}$. The pair becomes $10u + 2v = 4$ and $15u - 5v = -2$. Multiply the first by $5$ and the second by $2$, and add:

Then $2 + 2v = 4$ gives $v = 1$. So $x + y = 5$ and $x - y = 1$. Adding these gives $x = 3$, and $y = 2$.

Practise with new numbers

Try it in Python

The program works out $X$, $Y$ and $D$ for the worked pair, and then solves the reducible pair by changing letters.

Common mistakes

• Using $c$ with the wrong sign. For $3x + 4y = 10$, the constant is $c = -10$, not $10$.
• Mixing up the order in $X$. It is $b_1c_2 - b_2c_1$, the product down the main diagonal first.
• Stopping at $u$ and $v$. A reducible pair asks for $x$ and $y$, so take reciprocals at the end.
• Dividing by zero. If $D = 0$, the lines are parallel or the same line, and there is no single solution.

Remember these

• Write the pair as $ax + by + c = 0$, then $x = \dfrac{X}{D}$ and $y = \dfrac{Y}{D}$.
• $X = b_1c_2 - b_2c_1$, $Y = c_1a_2 - c_2a_1$ and $D = a_1b_2 - a_2b_1$.
• For a pair in $\dfrac{1}{x}$ and $\dfrac{1}{y}$, let $u = \dfrac{1}{x}$ and $v = \dfrac{1}{y}$, solve, and change back.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write $3x + 4y = 10$ in the form $ax + by + c = 0$. What is $c$?

2. For a pair, $D = 0$. Can cross-multiplication give a single solution?

3. Use cross-multiplication on $x + 2y = 7$ and $3x - y = 7$.

4. Which new letters make $\dfrac{6}{x} - \dfrac{2}{y} = 4$ and $\dfrac{3}{x} + \dfrac{1}{y} = 7$ linear?

5. After solving a reducible pair, Nisha finds $u = 4$ and $v = 5$, where $u = \dfrac{1}{x}$ and $v = \dfrac{1}{y}$. What are $x$ and $y$?