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This page sets out the method for simultaneous 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 Simultaneous equations topic page.
When both equations are in the form , line them up and cancel a letter.
When one equation starts '' or '', swap that expression into the other equation.
Solve 2x+y=7 and x+y=4 by subtracting to remove y.
Sometimes the mark is only for x.
After x is known, substitute to find y. Give only y when that is all they ask.
Adding 2x+3y to 4x+5y gives 6x+8y. This is the same as adding two equations when you eliminate.
Add x terms to x terms and y terms to y terms.
Put the pair back into both equations.
If the matching terms have the same sign, subtract the equations. If they have opposite signs, add them. Write that rule next to the working until it is automatic. x+y=10 and x-y=4 have opposite signs on y, so add: 2x=14, x=7, then y=3. Check in the second equation: 7-3=4. The check uses the equation you did not rearrange to find the second letter, so it is a real test rather than a copy of the last line.
Subtracting when you should add, or the other way round, does not just lose a mark. It produces a pair that fails the check. The check is there so you notice. 2y-(-2y) is 4y, not 0. A double negative in a subtraction is the line to write out in full.
Method: 1. Subtract the equations when the matching terms have the same sign. 2. Add the equations when those matching terms have opposite signs. 3. Keep that rule written beside the working until you can apply it without a reminder.
When no coefficients match, multiply one or both equations so that they do. Multiply every term, including the right-hand side. From 2x+y=8, multiplying by 3 gives 6x+3y=24. Leaving the 8 unmultiplied is a different equation and the rest of the solution is fiction.
Then eliminate as usual. With 4x+3y=18 and 6x+3y=24, the y terms have the same sign, so subtract: 2x=6, x=3. Substitute into the simpler original equation: 2× 3+y=8, so y=2. Check in the equation you have not just used: 4× 3+3× 2=12+6=18. Write "correct" next to a successful check. It is the difference between a probable answer and a confirmed one.
Method: 1. If no pair of coefficients matches, multiply one or both equations until a pair does. 2. Multiply every term, including the number on the right-hand side. 3. Multiplying 2x+y=8 by 3 gives 6x+3y=24.
Define the letters in a sentence before you write any equation. "Let t be the cost of a tea in pence, and c the cost of a cake." Two teas and one cake costing 540 pence is 2t+c=540. One tea and two cakes costing 630 pence is t+2c=630. The sentence that defines the letters is often the first mark.
Solve, then translate back into money. A solution of t=150 is £1.50, not 150 pounds. The units you chose at the start decide the units of the answer. Check both prices in both sentences. If one equation fails, the multiplication of a right-hand side is the first place to look.
If the story uses three of one item, expect to multiply. Do that multiplication on a separate labelled line, such as "(2) × 3", so the marker can see which equation grew. Then eliminate, substitute, and convert the pair back into the nouns of the question: a price, a number of tickets, an age. A correct pair written only as x=3, y=2, when the question asked for the cost of a cake, has stopped one sentence too soon. Name the cake price in a final line.
Method: 1. Define each letter in a sentence before you write an equation. 2. If t is the cost of a tea in pence and c is the cost of a cake, two teas and one cake is 2t+c=540. 3. One tea and two cakes costing 630 pence is t+2c=630.
A simultaneous equations question is usually worth three or four marks, and the marks follow the method. One mark is for making the coefficients of one letter match and eliminating it, one is for the value of the first letter, and one or two are for the second letter. If you work in your head and write only the answers, one slip can cost every mark. Write each line, and label the equations (1) and (2).
After you have found both letters, put your pair into the equation you did not use for the second letter, to check it. The check is quick and it catches almost every sign error. The final answer must show both values, for example x = 3 and y = 2. An answer that gives only x is half a solution and loses the last mark.
Method: 1. Label the equations (1) and (2) and write them one above the other. 2. Make the coefficients of one letter match, then add or subtract. 3. Solve for the first letter, then substitute to find the second. 4. Check the pair in the other equation and write both values clearly.
Tip: The check equation is the one you did not use to find the second letter.
Real questions rarely hand you two neat equations with the letters lined up. You might see 2y = 4x - 7 next to 2x + 4y = -9. First rearrange so each equation has the form ax + by = c. Move the x term across to get 4x - 2y = 7. Now the letters line up and elimination works. Take care with signs when you move terms.
If one equation is already y = ldots, use substitution. For x + y = 10 and y = 2x + 1, replace y in the first equation with 2x + 1 to get x + 2x + 1 = 10. This gives 3x = 9, so x = 3, and then y = 7. Substitution is quicker when one letter is already isolated, and elimination is quicker when both equations are in standard form.
Method: 1. Rearrange each equation into the form ax + by = c if you want to eliminate. 2. If a letter is already the subject, substitute it into the other equation. 3. Solve the one-letter equation you get. 4. Find the second letter and check in both original equations.
In a worded problem, define your letters first. Write "Let a be the price of an adult ticket in pounds and c the price of a child ticket in pounds." Then turn each sentence into one equation. Three adult tickets and two child tickets costing £57 becomes 3a + 2c = 57. Make sure the units are the same in both equations before you start, for example both in pence.
After solving, read the question again and answer exactly what it asks. It might want the cost of one adult ticket, or the total for a different set of tickets. Check the answers are sensible: prices must be positive, and a number of people must be a whole number. Finally, put your values back into both original sentences to be sure they work in the story.
Method: 1. Write "Let ..." for each unknown with its units. 2. Turn each sentence into an equation. 3. Solve by elimination or substitution. 4. Answer the actual question and check in the story.
Higher questions sometimes include a third letter, such as k, in the right-hand sides. For 3x + 2y = 13k and 2x - y = 4k you solve in exactly the same way, treating k like a number. Multiply the second equation by 2 to get 4x - 2y = 8k, add to get 7x = 21k, so x = 3k, then y = 2k. The answers are expressions in k, and that is correct.
Not every pair of equations has a single solution. If rearranging both into y = mx + c gives the same gradient but different intercepts, the lines are parallel and never meet, so there is no solution. If you reach a false statement such as 0 = 5 during elimination, that is the same sign. Say clearly that the lines are parallel, and give the gradient to justify it.
Method: 1. Treat the extra letter as a number and eliminate as usual. 2. Give each answer as an expression, not a number. 3. To test for no solution, write both equations as y = mx + c. 4. Same gradient and different intercepts means parallel lines and no solution.
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