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This page sets out the method for solving linear equations in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Solving linear equations topic page.
When you see an equation with one term, undo the add/subtract first, then the multiply/divide.
When you see brackets or on both sides, tidy first, then balance.
In 2x + 3 = 11, subtract 3, then divide by 2.
x = 4.
Whatever you do to one side, do to the other.
The equation stays balanced.
In 3x + 2 = x + 10, subtract x first so the letter is on one side.
Then undo the remaining add and multiply.
Expand a bracket before you collect, or divide both sides by the bracket’s factor if it is a common factor.
5(x + 1) = 20 can be divided by 5 first.
Substitute your value into the original equation.
Both sides should match.
If x/2 + 1 = 5, subtract 1, then multiply by 2.
The fraction bar is a division.
Read 4x-7=21 as "something was multiplied by 4, then 7 was subtracted". Undo the subtraction first: add 7 to both sides, so 4x=28. Then divide by 4, so x=7. Check: 4× 7-7=28-7=21. A common slip is to divide only the 4x and the 21 by 4. That gives x-7=5.25, which is wrong because the -7 was left untouched. Every term has to be divided if you divide a whole side.
The same order works when the second operation is addition. 3x+4=19 becomes 3x=15, then x=5. Write one operation on each line and line the equals signs up. A method mark sits on those lines. A correct answer with no working can still score, but a slip with no working cannot be rescued.
Method: 1. Read 4x-7=21 as a multiplication by 4, followed by a subtraction of 7. 2. Undo the subtraction first by adding 7 to both sides, which gives 4x=28. 3. Then divide both sides by 4 to get x=7.
Expand 2(x+6)=20 to 2x+12=20. Then it is an ordinary two-step equation: 2x=8, so x=4. Check inside the original bracket: 4+6=10 and 2× 10=20. Expanding 3(2x+5) as 6x+5 multiplies only the first term. Both terms inside the bracket are multiplied: 6x+15.
A single fraction such as (x+3)/2=7 is cleared by multiplying both sides by the denominator. x+3=14, so x=11. Multiplying only the top, or only one side, breaks the balance. If there are two fractions, multiplying through by a common denominator is the same idea. At this stage one denominator is enough, and the multiply happens before you expand or collect.
Method: 1. Expand 2(x+6)=20 to 2x+12=20 before you solve. 2. This is now a two-step equation: 2x=8, so x=4. 3. Substitute x=4 back into the original bracket: 2×(4+6)=20.
In 5(x-2)=3x+6 the letter appears in two places. Expand first: 5x-10=3x+6. Subtract 3x from both sides so the letter terms sit together: 2x-10=6. Add 10: 2x=16. Divide by 2: x=8. Check in the original: 5(8-2)=30 and 3× 8+6=30. The two sides match, so the solution is right.
Subtracting a term moves it with the opposite sign. If you "move" 3x and leave it as +3x on the other side, you have added instead of subtracted. Writing the operation in words beside the line, "subtract 3x", stops that. A negative solution is still checked the same way. If both sides of the check agree, keep the negative. Do not flip it to make the answer look friendlier.
Method: 1. In 5(x-2)=3x+6 the unknown appears on both sides. 2. Expand the bracket first, giving 5x-10=3x+6. 3. Subtract 3x from both sides so the unknown terms sit together: 2x-10=6.
When x appears on both sides, the plan is to collect all the x terms on one side and all the plain numbers on the other. In 9x-4=4x+21 subtract 4x from both sides to get 5x-4=21, then add 4 to get 5x=25 and divide to get x=5. Each line is the line above with the same operation done to both sides.
Choose the side that keeps the x term positive. Subtract the smaller x term from both sides, because a negative x term leads to sign errors. If you do end up with -x or a negative coefficient, such as 7-3x=2x-13, add 3x to both sides to get 7=5x-13 and carry on. Use one clear line for each move and always finish with a check in the original equation.
Method: 1. Decide which side has the larger x coefficient and subtract the smaller one from both sides. 2. Add or subtract the plain number so that the numbers are all on the other side. 3. Divide both sides by the number in front of x. 4. Check by substituting your answer into both sides of the original equation.
Tip: If both sides give the same number when you substitute, the solution is correct. If they differ, there is a slip.
Expand every bracket first, one at a time, and then treat what is left as an equation with letters on both sides. For 4(2x+3)=6(x-1)+5 the left side is 8x+12 and the right side is 6x-6+5=6x-1. Subtract 6x to get 2x+12=-1, subtract 12 to get 2x=-13 and divide to get x=-6.5.
Take care with minus signs in front of brackets. A term such as -3(x-4) expands to -3x+12 because -3×-4=+12. After expanding, tidy each side by collecting like terms before you start to move terms across. Exam questions frequently give a negative or decimal answer to check you have not made an error with signs, so do not be alarmed by an answer like -6.5.
Method: 1. Multiply every term in each bracket by the number outside it. 2. Collect like terms on each side separately. 3. Move the x terms to one side and the numbers to the other. 4. Divide, then check by substituting into the original equation.
To remove a fraction, multiply every term on both sides by the denominator. For (x+5)/3=4 multiply both sides by 3 to get x+5=12, so x=7. If there is more than one fraction, multiply by a number that every denominator divides into. For x/4+7=x/3+5 use 12, giving 3x+84=4x+60 and so x=24.
Multiply every term, not just one. In x/2+3=7 the 3 must also be multiplied by 2, which gives x+6=14. A shorter route is to subtract 3 first, then multiply by 2, because the fraction is on its own. Both methods are valid. When the unknown is inside a bracket on top, such as (2x-1)/3, multiplying by 3 leaves the bracket as it is.
Method: 1. Find a number that every denominator divides into. 2. Multiply every term on both sides by that number. 3. Cancel the fractions and write out the new equation without them. 4. Solve the new equation, then check in the original.
Tip: Multiply the whole of each side. Writing brackets around both sides, such as 12(x/4+7), helps you not to miss a term.
Many questions give a value and ask whether it is the solution. Put the number into both sides. For 5x-3=2x+9 with x=4, the left side is 17 and the right side is 17, so it is correct. Writing both values and a sentence is what earns the mark. A guess or "it works" with no numbers does not.
Another style gives someone's working and asks you to find the error. Read down the lines until the equation stops being equivalent. For 5(x-3)=2x+18 a student wrote 5x-3=2x+18. The 5 was only applied to x, not to -3. Say that clearly, and then solve the equation yourself: 5x-15=2x+18 gives 3x=33 and x=11.
Method: 1. Substitute the number into the left side and the right side separately. 2. State both values and say whether they are equal. 3. In a worked solution, find the first line that does not follow from the line before. 4. Describe the error in words and then solve the equation correctly.
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