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Systems of Linear Equations: math lesson, Grade 8 – download the PDF

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Math lessons Grade 8 : Systems of Linear Equations — Zyro the alien explorer of Planète Maths

Two lines on a map can cross, run side by side forever, or sit exactly on top of each other. In this chapter you will learn to answer a big question about two linear equations at once: which pair of numbers makes both of them true? You will find it with a graph, with substitution, and with elimination, and you will use it to crack word problems.

1. What is a system of equations?

A single equation such as \( y = 2x + 1 \) has infinitely many solutions: every point of its line. When two equations must be true at the same time, we only keep the pairs that work in both.

Definition

A system of linear equations is a set of two (or more) linear equations that use the same variables. A solution of the system is an ordered pair \( (x, y) \) that makes every equation true.

Example 1: checking a solution

Is \( (3, 2) \) a solution of \( \begin{cases} x + y = 5 \\ 4x - y = 10 \end{cases} \)?

First equation: \( 3 + 2 = 5 \), true. Second equation: \( 4(3) - 2 = 10 \), true. Both work, so \( (3, 2) \) is the solution.

A pair that works in only one equation is not a solution of the system. Always test both equations.

2. Solving by graphing

Each equation is a line in the coordinate plane. A point on the first line satisfies the first equation, and a point on the second line satisfies the second one. A point on both lines satisfies both equations: it is the intersection point.

Method: graphing

  1. Write each equation in slope-intercept form \( y = mx + b \).
  2. Graph each line using the y-intercept \( b \) and the slope \( m \).
  3. Read the coordinates of the point where the lines cross.
  4. Check the point in both original equations.
Example 2: graphing

Solve \( y = x + 1 \) and \( y = -2x + 7 \) by graphing.

The purple line starts at \( 1 \) on the y-axis and goes up 1 for each step right. The orange line starts at \( 7 \) and goes down 2 for each step right. They cross at \( (2, 3) \). Check: \( 3 = 2 + 1 \) and \( 3 = -2(2) + 7 \). Both are true.

-1123456-1123456789(2, 3)

The two lines cross at (2, 3).

Watch out

Graphing gives exact answers only when the crossing point falls on a grid corner. If the lines cross between grid lines, your reading is only an estimate, so switch to an algebraic method.

3. Solving by substitution

Substitution works best when one equation already says y = something (or x = something). You replace that variable in the other equation, which leaves an equation with only one unknown.

Method: substitution

  1. Isolate one variable in one of the equations.
  2. Substitute that expression into the other equation.
  3. Solve for the remaining variable.
  4. Substitute back to find the other variable, then check in both equations.
Example 3: substitution

Solve \( \begin{cases} y = 3x - 4 \\ 2x + y = 11 \end{cases} \).

Replace \( y \) in the second equation: \( 2x + (3x - 4) = 11 \), so \( 5x - 4 = 11 \), so \( 5x = 15 \) and \( x = 3 \). Then \( y = 3(3) - 4 = 5 \). The solution is \( (3, 5) \). Check: \( 2(3) + 5 = 11 \). True.

4. Solving by elimination

Elimination makes one variable disappear by adding or subtracting the equations. It is powerful when both equations are in standard form \( Ax + By = C \).

Method: elimination

  1. Line up the equations as \( Ax + By = C \).
  2. If needed, multiply one or both equations so that one variable has opposite coefficients (like \( 4y \) and \( -4y \)).
  3. Add the equations. One variable vanishes.
  4. Solve, substitute back, and check.
Example 4: elimination with a multiplier

Solve \( \begin{cases} 3x + 2y = 16 \\ x - y = 2 \end{cases} \).

Multiply the second equation by 2: \( 2x - 2y = 4 \). Add it to the first: \( 5x = 20 \), so \( x = 4 \). Then \( 4 - y = 2 \) gives \( y = 2 \). The solution is \( (4, 2) \). Check: \( 3(4) + 2(2) = 16 \). True.

Zyro’s tip

On my planet we choose the method that looks shortest: if a variable is already alone, substitute; if the equations are in standard form, eliminate. Both give the same answer, so pick the quicker road!

5. How many solutions?

Two lines in a plane can be in exactly three positions. That gives exactly three possible outcomes.

Lines Slopes and intercepts Number of solutions
Cross at one point Different slopes Exactly one
Parallel, never meet Same slope, different y-intercepts None
Same line (coincident) Same slope, same y-intercept Infinitely many
Property

A system of two linear equations has either exactly one solution, no solution, or infinitely many solutions. It can never have exactly two.

6. Parallel and coincident lines

When you solve algebraically, the special cases show up as a statement that does not involve any variable.

  • A false statement such as \( 0 = 5 \) means the lines are parallel: no solution.
  • A always true statement such as \( 6 = 6 \) means the lines are the same: infinitely many solutions.

-3-2-11234-6-4-22468

Parallel lines: same slope 2, different y-intercepts, no crossing point.

Example 5: no solution

Solve \( y = 2x + 1 \) and \( y = 2x - 3 \). Substitute: \( 2x + 1 = 2x - 3 \), so \( 1 = -3 \). That is false, so there is no solution. Both slopes are 2 but the intercepts differ.

-4-3-2-11234-2-1123456

Coincident lines: the dashed line sits exactly on the purple one.

Example 6: infinitely many solutions

Solve \( y = x + 2 \) and \( 2y = 2x + 4 \). Divide the second equation by 2: \( y = x + 2 \). It is the same line, so every point of it is a solution. Substituting gives \( 2(x + 2) = 2x + 4 \), which simplifies to \( 4 = 4 \), always true.

7. Word problems with systems

Many real situations have two unknowns and two facts. Translate each fact into an equation.

Method: word problems

  1. Define two variables with units, for example \( a \) = number of adult tickets.
  2. Write one equation for each fact (a total count, a total cost, a distance...).
  3. Solve the system with the best method.
  4. Answer in a full sentence with units, and check against the story.
Example 7: a snack stand

A stand sells wraps for $6 and juices for $2. Today it sold 30 items and earned $108. How many wraps were sold?

Let \( w \) be wraps and \( j \) juices. Then \( w + j = 30 \) and \( 6w + 2j = 108 \). From the first, \( j = 30 - w \). So \( 6w + 2(30 - w) = 108 \), \( 4w + 60 = 108 \), \( w = 12 \), and \( j = 18 \). Check: \( 72 + 36 = 108 \). The stand sold 12 wraps.

Graphs help to compare options. In the figure below, two plans cost the same after 10 rides; before that, plan B is cheaper, and after that plan A is cheaper.

24681012141020304050(10, 35)

Cost in dollars against number of rides: the plans meet at (10, 35).

Key takeaways

  • A solution of a system is an ordered pair that makes all equations true.
  • Graphing: the solution is the intersection point of the lines. Check it in both equations.
  • Substitution: isolate a variable, replace it in the other equation, solve, then back-substitute.
  • Elimination: create opposite coefficients, add the equations, solve, then back-substitute.
  • Different slopes give one solution; same slope with different intercepts gives none; identical lines give infinitely many.
  • A false statement like \( 1 = -3 \) means no solution; a true one like \( 4 = 4 \) means infinitely many.
  • In word problems, define the variables, write two equations, solve, and answer with units.
Do the practice problems : Systems of Linear Equations: math lesson, Grade 8 – Planète MathsTake the quiz : Systems of Linear Equations: math lesson, Grade 8 – Planète Maths

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