Matematica

Systems of Equations: Solved Exercises With Substitution and Elimination

How to solve a system of two linear equations using the substitution method and the elimination method, step by step, plus five solved exercises of increasing difficulty with verified answers.

Recommended for: Grade 9 ยท Grade 10

A system of two linear equations in two unknowns is solved by finding the pair (x, y) that satisfies both equations at the same time. The two main methods are substitution and elimination: they lead to the same result, only the path is different.

Substitution or elimination?

With substitution you solve one equation for a variable and put it in place of that variable in the other equation, leaving a single equation in a single unknown. It works best when one of the two equations already has a variable isolated or with coefficient 1.

With elimination you add or subtract the two equations, after multiplying them by a number if needed, so that one variable disappears. It is the faster route when the coefficients can be made opposite.

One check at the end

Once you have the pair (x, y), substitute it into both original equations: both must come out true. The exercises below start with systems where a variable is already isolated and end with a system that needs both equations multiplied before eliminating. Every answer is verified by substituting the values found.

Solved exercises

1. Solve the system: x + y = 5 ; y = x + 1 base

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  1. The second equation already gives y on its own: y = x + 1.
  2. Substitute into the first: x + (x + 1) = 5.
  3. Simplify: 2x + 1 = 5, so 2x = 4 and x = 2.
  4. Find y: y = x + 1 = 2 + 1 = 3.

Answer: x = 2 ; y = 3

2. Solve the system: x + y = 10 ; x - y = 4 base

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  1. Use elimination: add the two equations term by term.
  2. (x + y) + (x - y) = 10 + 4, that is 2x = 14.
  3. So x = 7.
  4. Substitute into the first: 7 + y = 10, so y = 3.

Answer: x = 7 ; y = 3

3. Solve the system: y = x - 4 ; 2x + y = 11 intermedio

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  1. The first gives y = x - 4: substitute it into the second.
  2. 2x + (x - 4) = 11, that is 3x - 4 = 11.
  3. 3x = 15, so x = 5.
  4. Find y: y = x - 4 = 5 - 4 = 1.

Answer: x = 5 ; y = 1

4. Solve the system: 2x + 3y = 12 ; x - y = 1 intermedio

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  1. Use elimination: multiply the second equation by 3.
  2. This gives 3x - 3y = 3.
  3. Add it to the first: (2x + 3y) + (3x - 3y) = 12 + 3, that is 5x = 15.
  4. x = 3; substitute into the second: 3 - y = 1, so y = 2.

Answer: x = 3 ; y = 2

5. Solve the system: 3x + 2y = 4 ; 2x - 3y = 7 avanzato

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  1. Eliminate y: multiply the first by 3 and the second by 2.
  2. 9x + 6y = 12 and 4x - 6y = 14.
  3. Add: 13x = 26, so x = 2.
  4. Substitute into the first: 3(2) + 2y = 4, that is 6 + 2y = 4, 2y = -2, y = -1.

Answer: x = 2 ; y = -1

FAQ

Should I use substitution or elimination?

It depends on the system. Substitution is best when one variable is already isolated or has coefficient 1; elimination is faster when the coefficients can be made opposite by adding or subtracting the equations.

What does it mean for a system to have no solution or infinitely many?

A system has no solution when no pair (x, y) satisfies both equations (parallel lines); it has infinitely many solutions when the two equations describe the same line.

How can I be sure the solution is correct?

Substitute the values of x and y back into both original equations: both must come out true. If even one fails, there is a mistake in the calculations.