This note covers solving linear equations (including those with fractions), solving pairs of simultaneous equations, and rearranging formulae to change the subject. These methods are used any time a real situation is turned into algebra, from ticket prices to physics formulae. Solving a linear and a non-linear pair together is an Extended skill.
Key ideas
Linear equations. An equation is a balance: whatever you do to one side, you do to the other. The aim is to get the unknown on its own.
- Expand any brackets first, then collect the unknown terms on one side and the numbers on the other.
- If there are fractions, multiply every term by a common multiple of the denominators to clear them.
- Check your answer by substituting it into the original equation.
- To form an equation from a word problem, let a letter stand for the unknown, then write down the information as an equation.
Simultaneous equations. Two equations with two unknowns can be solved together. There are two standard methods.
- Elimination. Multiply one or both equations so that one unknown has the same coefficient in both. If the signs are the same, subtract the equations. If the signs are opposite, add them. Solve for the remaining unknown, then substitute back.
- Substitution. Make one unknown the subject of one equation and substitute it into the other. This is the best choice when one equation is already in the form \(y=\ldots\), and it is the method for a linear and a non-linear equation (Extended).
Graphically, the solution is the point where the two lines cross. Always substitute your answers into both original equations as a check.
Rearranging formulae. Changing the subject means getting a chosen letter alone on one side. Undo operations in reverse order, as in solving an equation. If the subject appears twice, collect all terms containing it on one side, factorise, then divide. A formula with a power of the subject needs a root at the end, and the root may need a \(\pm\) if the context allows a negative value.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Linear equation | An equation in which the unknown appears only to the power 1, such as \(3x+5=17\). |
| Inverse operations | Addition and subtraction, multiplication and division, squaring and square roots. Use them to undo steps. |
| Simultaneous equations | Two equations that must both be true at the same time. |
| Elimination | Add or subtract equations to remove one unknown. |
| Substitution | Replace one unknown by an expression from the other equation. |
| Subject of a formula | The letter that stands alone on one side, such as \(V\) in \(V=IR\). |
| \(ax+b=c\Rightarrow x=\dfrac{c-b}{a}\) | General solution of a basic linear equation, \(a\ne0\). |
Worked examples
Example 1 Solve \(\dfrac{2x-1}{3}+\dfrac{x+4}{2}=4\).
The common multiple of 3 and 2 is 6. Multiply every term by 6:
\(6\times\dfrac{2x-1}{3}+6\times\dfrac{x+4}{2}=6\times4\), so \(2(2x-1)+3(x+4)=24\).
Expand the brackets: \(4x-2+3x+12=24\), which gives \(7x+10=24\).
Subtract 10: \(7x=14\). Divide by 7: \(x=2\).
Check: \(\dfrac{3}{3}+\dfrac{6}{2}=1+3=4\). Correct.
Answer: \(x=2\)
Example 2 (a) Three adult tickets and two child tickets for a museum cost £16. Five adult tickets and three child tickets cost £26. Find the price of each type. (b) (Extended) Solve \(y=x+1\) and \(x^2+y^2=25\) together.
(a) Let an adult ticket cost \(a\) pounds and a child ticket \(c\) pounds. Then \(3a+2c=16\) and \(5a+3c=26\).
Make the coefficients of \(c\) equal to 6: multiply the first equation by 3 and the second by 2.
\(9a+6c=48\) and \(10a+6c=52\).
The signs of \(6c\) are the same, so subtract: \(10a-9a=52-48\), giving \(a=4\).
Substitute into \(3a+2c=16\): \(12+2c=16\), so \(c=2\).
Check in the second equation: \(5\times4+3\times2=26\). Correct.
(b) Substitute \(y=x+1\) into the circle equation: \(x^2+(x+1)^2=25\).
Expand: \(x^2+x^2+2x+1=25\), so \(2x^2+2x-24=0\). Divide by 2: \(x^2+x-12=0\).
Factorise: \((x+4)(x-3)=0\), so \(x=-4\) or \(x=3\).
Use \(y=x+1\) to find each \(y\): \(x=3\) gives \(y=4\), and \(x=-4\) gives \(y=-3\).
Answer: (a) adult £4, child £2; (b) \((3,4)\) and \((-4,-3)\)
Example 3 (a) Make \(r\) the subject of \(V=\dfrac{1}{3}\pi r^2h\). (b) Make \(x\) the subject of \(y=\dfrac{2x+1}{x-3}\).
(a) Multiply both sides by 3: \(3V=\pi r^2h\). Divide by \(\pi h\): \(r^2=\dfrac{3V}{\pi h}\).
Take the square root. A radius is positive, so \(r=\sqrt{\dfrac{3V}{\pi h}}\).
(b) Multiply both sides by \((x-3)\): \(y(x-3)=2x+1\). Expand: \(xy-3y=2x+1\).
Collect the \(x\) terms on one side and everything else on the other: \(xy-2x=1+3y\).
Factorise the left side: \(x(y-2)=1+3y\). Divide by \((y-2)\): \(x=\dfrac{3y+1}{y-2}\).
Check with \(x=5\): \(y=\dfrac{11}{2}=5.5\), and \(\dfrac{3(5.5)+1}{5.5-2}=\dfrac{17.5}{3.5}=5\). Correct.
Answer: (a) \(r=\sqrt{\dfrac{3V}{\pi h}}\); (b) \(x=\dfrac{3y+1}{y-2}\)
Common errors
- Multiplying only some terms by the common multiple when clearing fractions. Every term, including the number on the other side, must be multiplied.
- Sign mistakes when expanding a bracket with a minus in front, such as \(-(x-2)=-x+2\), not \(-x-2\).
- In elimination, subtracting when the signs are opposite (or adding when they are the same), which leaves the unknown in place.
- Finding only one unknown and stopping. A pair of simultaneous equations needs both values, and each should be checked in both equations.
- When the subject appears twice, trying to divide before collecting the terms. Collect, factorise, then divide.
Exam tip: Label your equations (1) and (2) and write down what you multiply by. If you make a slip, the working shows exactly where, and you can fix it in seconds.
Check yourself
- Solve \(5(x-2)=3x+8\).
Answer
\(5x-10=3x+8\), so \(2x=18\) and \(x=9\). - Solve \(2x+3y=7\) and \(4x-3y=5\).
Answer
The \(y\) terms have opposite signs, so add: \(6x=12\), \(x=2\). Then \(4+3y=7\), so \(y=1\). Check: \(8-3=5\). Correct. - Make \(m\) the subject of \(E=\dfrac12mv^2\).
Answer
Multiply by 2: \(2E=mv^2\). Divide by \(v^2\): \(m=\dfrac{2E}{v^2}\). - (Extended) Solve \(y=2x\) and \(x^2+y=15\).
Answer
Substitute: \(x^2+2x=15\), so \(x^2+2x-15=0\) and \((x+5)(x-3)=0\). Then \(x=-5\), \(y=-10\), or \(x=3\), \(y=6\).