How to solve simultaneous linear equations
A solution to two simultaneous equations must satisfy both equations at the same time. For two linear equations in x and y, elimination and substitution let you reduce the pair to an equation with one unknown. Write the steps clearly so you can check the operation that produced each new line.
Choose a method from the coefficients
Line up the x terms, y terms and constants, then label the equations A and B. If one variable already has matching coefficients, elimination may be a short route. Subtract equations with equal coefficients, or add equations with opposite coefficients, to remove that variable.
If a variable is already isolated, or has a coefficient of 1 or −1, substitution may be convenient. Rearrange one equation to express that variable in terms of the other, then replace it in the second equation. Either method must preserve the original relationships.
When multiplying an equation, multiply every term on both sides. When subtracting equations, subtract the whole second expression, including negative terms. Keep parentheses until you have applied the signs.
Eliminate one variable in a worked example
Use this invented practice pair: A is 2x + 3y = 19, and B is 3x + 2y = 16. Neither variable has equal coefficients yet. To eliminate x, multiply A by 3 and B by 2. The new equations are 6x + 9y = 57 and 6x + 4y = 32.
Subtract the second new equation from the first: (6x + 9y) − (6x + 4y) = 57 − 32. The x terms cancel, leaving 5y = 25, so y = 5. Notice that the constants were multiplied too: keeping 19 and 16 on the right would have changed the equations.
Put y = 5 into original equation A: 2x + 3(5) = 19. This gives 2x + 15 = 19, then 2x = 4 and x = 2. Write the answer as x = 2, y = 5, with units if the variables represent quantities in a word problem.
Use substitution when a variable is easy to isolate
For a separate practice pair, take x + y = 7 and 2x + 3y = 19. The first equation gives x = 7 − y. Replace x in the second equation with that entire expression: 2(7 − y) + 3y = 19.
Expand to get 14 − 2y + 3y = 19, so 14 + y = 19 and y = 5. Then x = 7 − 5 = 2. The parentheses matter: 2(7 − y) becomes 14 − 2y, not 14 − y.
If substitution produces fractions, keep them exact while solving. You can multiply both sides of the resulting equation by a common denominator to remove them. Follow any answer-format instruction in the question before rounding a final value.
Check both original equations
For the elimination example, check A: 2(2) + 3(5) = 4 + 15 = 19. Check B: 3(2) + 2(5) = 6 + 10 = 16. Both match their original right-hand sides. Checking only the equation used to find x would leave the other relationship untested.
If a check fails, return to the first line that no longer follows from the preceding equations. Inspect whole-equation multiplication, subtraction signs, bracket expansion and the substitution used to recover the second variable. Correct that step before repeating later calculations.
Use this working template: “Original equations: [A and B]. Method: [variable to eliminate or expression to substitute]. One-variable equation: [result]. First value: [value]. Other value: [substitution into an original equation]. Checks: [both original equations].”
Recognise when there is no single solution
If elimination gives a contradiction such as 0 = 3, no pair of values satisfies both equations. For example, x + y = 4 and 2x + 2y = 11 would imply both 2x + 2y = 8 and 2x + 2y = 11.
If the two equations describe the same line, there are infinitely many solutions. For example, x + y = 4 and 2x + 2y = 8 reduce to the same relationship; subtraction gives 0 = 0. You still need x + y = 4, but the pair does not select one value of x and one value of y.
These examples concern two linear equations in two variables. If your equations contain squares, products of variables or other nonlinear terms, the number of solutions and the method may differ. Start by checking what kind of equations the question actually gives.
