This note covers solving linear equations and simultaneous equations, including equations with fractions, forming equations from word problems, and (at Higher tier) pairs of equations where one is quadratic. These methods are used to find unknown values in problems about cost, speed, geometry and design.
Key ideas
Linear equations. An equation is a balance: whatever you do to one side, you must do to the other. Aim to get the unknown on its own. The usual steps are to expand brackets, collect the unknown terms on one side and the numbers on the other, then divide.
For equations with fractions, multiply every term on both sides by the common denominator to remove the fractions. If the unknown is on both sides, subtract the smaller unknown term from both sides so the coefficient stays positive. Always check your answer by substituting it back into the original equation.
Forming equations. In a word problem, let a letter stand for the unknown, write expressions for the other quantities, and then write a statement of equality from the problem. Perimeters, ages and costs often give such statements.
Simultaneous equations. When there are two unknowns, you need two equations. The solution is the pair of values that satisfies both. Graphically, it is the point where the two lines cross.
- Elimination. Multiply one or both equations so that the coefficient of one unknown is the same in both. Then add the equations if the signs are different, or subtract them if the signs are the same. Solve for one unknown, then substitute to find the other.
- Substitution. Rearrange one equation to make an unknown the subject, then replace it in the other equation. This method is natural when one equation already has the form \(y=\ldots\) or \(x=\ldots\).
One linear and one quadratic equation (Higher tier). Use substitution. Replace one unknown in the quadratic equation with the linear expression. This gives a quadratic equation in one unknown, which you solve (usually by factorising or the formula). Then substitute each solution back into the linear equation to find the matching value of the other unknown. There are usually two pairs of solutions, which correspond to the two points where a line meets a curve.
Always check the final pairs in both of the original equations.
Choosing a method. Use elimination when both equations have the form \(ax+by=c\). Use substitution when one equation is already written with \(x\) or \(y\) as the subject. Both methods give the same answer, so choose the one with less arithmetic.
Linking to graphs. Each linear equation in two unknowns represents a straight line. If the lines cross, the coordinates of the crossing point are the solution. This is useful for checking an answer and for reading approximate solutions from a sketch. For a line and a curve, the number of crossing points tells you how many solution pairs to expect.
Inequalities and equations. Do not confuse equations with inequalities. An equation has one or two exact solutions here, but an inequality has a range of values. Solving inequalities is covered in the note on quadratic equations and inequalities.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Linear equation | An equation where the unknown has power 1 only, such as \(3x+5=17\). |
| Coefficient | The number multiplying an unknown, such as 3 in \(3x\). |
| Simultaneous equations | Two (or more) equations that are true for the same values of the unknowns. |
| Elimination | Make coefficients match, then add or subtract to remove one unknown. |
| Substitution | Replace one unknown with an expression from the other equation. |
| Solution of simultaneous equations | Intersection point of the two graphs. |
| Linear and quadratic pair | Substitute the linear expression into the quadratic, solve, then find the other unknown (Higher tier). |
Worked examples
Example 1 Solve \(\dfrac{2x+1}{3}-\dfrac{x-2}{4}=2\).
The common denominator of 3 and 4 is 12. Multiply every term by 12: \(12\times\dfrac{2x+1}{3}-12\times\dfrac{x-2}{4}=12\times2\).
This gives \(4(2x+1)-3(x-2)=24\).
Expand the brackets, taking care with the minus sign: \(8x+4-3x+6=24\).
Collect like terms: \(5x+10=24\), so \(5x=14\).
Divide: \(x=\dfrac{14}{5}=2.8\).
Check: \(\dfrac{2(2.8)+1}{3}=\dfrac{6.6}{3}=2.2\) and \(\dfrac{2.8-2}{4}=0.2\). Then \(2.2-0.2=2\). Correct.
Answer: \(x=2.8\).
Example 2 At a theme park, 3 adult tickets and 2 child tickets cost £39. 2 adult tickets and 5 child tickets cost £48. Find the price of each type of ticket.
Let an adult ticket cost \(a\) pounds and a child ticket cost \(c\) pounds.
Equation 1: \(3a+2c=39\). Equation 2: \(2a+5c=48\).
Make the coefficients of \(a\) equal. Multiply equation 1 by 2: \(6a+4c=78\). Multiply equation 2 by 3: \(6a+15c=144\).
The signs of \(6a\) are the same, so subtract the first from the second: \(11c=66\), so \(c=6\).
Substitute into equation 1: \(3a+12=39\), so \(3a=27\) and \(a=9\).
Check in equation 2: \(2(9)+5(6)=18+30=48\). Correct.
Answer: an adult ticket costs £9 and a child ticket costs £6.
Example 3 (Higher) Solve the simultaneous equations \(y=2x-1\) and \(y=x^2-4\).
Both expressions equal \(y\), so set them equal: \(x^2-4=2x-1\).
Rearrange to zero: \(x^2-2x-3=0\).
Factorise: \((x-3)(x+1)=0\), so \(x=3\) or \(x=-1\).
Use the linear equation to find \(y\). If \(x=3\): \(y=2(3)-1=5\). If \(x=-1\): \(y=2(-1)-1=-3\).
Check in \(y=x^2-4\): \(3^2-4=5\) and \((-1)^2-4=-3\). Both correct.
Answer: \(x=3,\ y=5\) and \(x=-1,\ y=-3\).
Common errors
- Changing only one side of an equation, or forgetting to multiply every term (including the number on the right) by the common denominator.
- Sign errors when subtracting equations: subtracting \(-4y\) means adding \(4y\).
- Finding \(x\) and forgetting to find \(y\), or giving the answer without pairing the values correctly.
- In the linear and quadratic case, giving only one solution pair, or substituting back into the quadratic and creating extra incorrect pairs. Use the linear equation to pair the values.
- Not checking in both original equations, so an arithmetic slip goes unnoticed.
Exam tip: write "Equation 1" and "Equation 2" and show the multiplication of each equation. Then check your solution in the equation you did not use for the final substitution.
Check yourself
- Solve \(5x-7=2x+11\).
Answer
\(3x=18\), so \(x=6\). Check: \(30-7=23\) and \(12+11=23\). - Solve \(3(2x-1)=4(x+5)\).
Answer
\(6x-3=4x+20\), so \(2x=23\) and \(x=11.5\). - Solve \(3x+2y=13\) and \(5x-4y=7\).
Answer
Multiply the first by 2: \(6x+4y=26\). Add to the second: \(11x=33\), so \(x=3\). Then \(9+2y=13\), so \(y=2\). - (Higher) Solve \(y=x+1\) and \(x^2+y^2=25\).
Answer
Substitute: \(x^2+(x+1)^2=25\), so \(2x^2+2x+1=25\) and \(x^2+x-12=0\). Factorise: \((x+4)(x-3)=0\). If \(x=3\), \(y=4\). If \(x=-4\), \(y=-3\).