Choose a Method, Count Solutions and Fix Errors | Grade 8 Math | LevelKite

In this lesson, $x$ and $y$ stand for real numbers. A system of linear equations is a pair of linear equations that use the same variables. A solution of the system is an ordered pair $(x, y)$ that makes both equations true at the same time. Each linear equation in $x$ and $y$ describes a line, the set of all points that make it true, so a solution is a point that lies on both lines.

You have already solved systems by substitution and by elimination. Now you will choose the method that fits a system, find how many solutions a system has, and find and fix errors in worked solutions.

Both methods work on any system of two linear equations, so choose the one that takes less work.

Substitution means solving one equation for a variable and putting that expression into the other equation. It is efficient when a variable is already alone, or when it has a coefficient (the number multiplying it) of $1$ or $-1$ , so you can isolate it just by adding or subtracting terms on both sides.
Elimination means adding or subtracting the equations so that one variable cancels. Opposite coefficients cancel when you add, and equal coefficients cancel when you subtract. It is efficient when a variable already has equal or opposite coefficients, or when multiplying a whole equation by a nonzero number gives it equal or opposite coefficients. Multiplying both sides of an equation by the same nonzero number keeps its solutions.

Whichever method you use, check your answer in both original equations.

Spotting the efficient method

System	What you notice	Efficient method
$3x + 2y = 16$ and $x - 4y = -4$	$x$ has coefficient $1$ in the second equation	Substitution
$2x + 3y = 13$ and $5x - 3y = 1$	The $y$ -coefficients, $3$ and $-3$ , are opposites	Elimination: add the equations
$2x - 3y = 4$ and $-4x + 6y = 7$	Multiplying the first equation by $2$ gives $4x$ , the opposite of $-4x$	Elimination after multiplying
$3x + 2y = 20$ and $x - y = 5$	$x$ has coefficient $1$ , and multiplying the second equation by $2$ gives $-2y$ , the opposite of $2y$	Either method

Worked example

Example 1: Choose a method and solve

Solve the system $3x + 2y = 16$ and $x - 4y = -4$ . Choose an efficient method and check your answer.

Choose substitution, because x has coefficient 1 in the second equation. Add 4y to both sides of that equation. Doing the same operation to both sides keeps its solutions.
$x - 4y + 4y = -4 + 4y \;\Rightarrow\; x = 4y - 4$
In every solution of the system, x equals 4y - 4, so replace x with 4y - 4 in the other equation, 3x + 2y = 16. This combines the two different equations.
$3(4y - 4) + 2y = 16$
Distribute the 3, then combine the like terms 12y and 2y. Each rewrite gives an equivalent expression, so the solutions do not change.
$12y - 12 + 2y = 16 \;\Rightarrow\; 14y - 12 = 16$
Add 12 to both sides. Then divide both sides by 14, which is not zero.
$14y = 28 \;\Rightarrow\; y = \tfrac{28}{14} = 2$
Substitute y = 2 into x = 4y - 4 to find x.
$x = 4(2) - 4 = 8 - 4 = 4$
Check the pair (4, 2) in both original equations. Both statements are true, so the pair is a solution.
$3(4) + 2(2) = 12 + 4 = 16 \quad \text{and} \quad 4 - 4(2) = 4 - 8 = -4$

Answer

The system has exactly one solution, $(4, 2)$ : $x = 4$ and $y = 2$ . The two lines cross at exactly one point.

Because each equation describes a line, the number of solutions matches how the two lines meet.

The lines cross at exactly one point: the system has exactly one solution.
The lines are parallel and distinct, meaning different lines with the same slope, so they never meet: the system has no solution.
The two equations describe the same line, so every point on it lies on both lines: the system has infinitely many solutions.

A sketch cannot settle which case you have, because lines that look parallel may still cross far off the page. Algebra gives an exact answer.

Key concept

Combine the two different equations, then read the result

Combine the two different equations: substitute an expression taken from one equation into the other equation, or add or subtract the equations or nonzero multiples of them. Then read what is left.

One value for each variable, and the pair checks in both original equations: exactly one solution. The lines cross at one point.
Both variables drop out and a true statement remains, such as $0 = 0$ : infinitely many solutions. The equations describe the same line.
Both variables drop out and a false statement remains, such as $0 = 5$ : no solution. The lines are parallel and distinct.

These readings hold only when the work combined the two different equations.

Worked example

Example 2: Count the solutions

How many solutions does the system $2x - 3y = 4$ and $-4x + 6y = 7$ have?

Choose elimination. No coefficient is 1 or -1, but multiplying the whole first equation by 2 gives the x-term 4x, the opposite of -4x. Multiplying both sides by the same nonzero number keeps the solutions.
$2(2x - 3y) = 2(4) \;\Rightarrow\; 4x - 6y = 8$
Add this equation to the second equation. Since -4x + 6y equals 7, adding them to the two sides of 4x - 6y = 8 keeps every shared solution. This combines the two different equations.
$(4x - 6y) + (-4x + 6y) = 8 + 7$
Combine like terms. The x-terms and the y-terms are opposites, so both variables drop out.
$(4x - 4x) + (-6y + 6y) = 15 \;\Rightarrow\; 0 = 15$
The statement 0 = 15 is false for every x and y, so it is a contradiction. It came from adding nonzero multiples of the two different equations, so no ordered pair makes both equations true.
Connect this to the lines. Solve each original equation for y: subtract 2x from both sides of the first and divide both sides by -3; add 4x to both sides of the second and divide both sides by 6. Neither divisor is zero.
$-3y = -2x + 4 \Rightarrow y = \tfrac{2}{3}x - \tfrac{4}{3} \qquad 6y = 4x + 7 \Rightarrow y = \tfrac{2}{3}x + \tfrac{7}{6}$

Answer

The system has no solution. Both lines have slope $\tfrac{2}{3}$ but different $y$ -intercepts, $-\tfrac{4}{3}$ and $\tfrac{7}{6}$ , so they are parallel and distinct and never meet.

Worked example

Example 3, mixed review: Choose, solve and count

Choose a method, solve the system $-x + 2y = 3$ and $3x - 6y = -9$ , and state the number of solutions.

Choose substitution, because x has coefficient -1 in the first equation. (Elimination would also work after multiplying the first equation by 3.) Add x to both sides, then subtract 3 from both sides.
$-x + 2y = 3 \;\Rightarrow\; 2y = 3 + x \;\Rightarrow\; x = 2y - 3$
Substitute 2y - 3 for x in the other equation, 3x - 6y = -9.
$3(2y - 3) - 6y = -9$
Distribute the 3 and combine like terms. Both variables drop out.
$6y - 9 - 6y = -9 \;\Rightarrow\; -9 = -9$
The statement -9 = -9 is true for every value of y, so it is an identity. It came from substituting an expression from the first equation into the other equation, so every pair (2y - 3, y) makes both equations true. Multiplying the first equation by -3 gives the second, so both describe the same line.
$-3(-x + 2y) = -3(3) \;\Rightarrow\; 3x - 6y = -9$
Check two of the solutions in both original equations: y = 0 gives (-3, 0), and y = 2 gives (1, 2).
$-(-3) + 2(0) = 3,\; 3(-3) - 6(0) = -9 \qquad -(1) + 2(2) = 3,\; 3(1) - 6(2) = -9$

Answer

The system has infinitely many solutions: every ordered pair on the line $-x + 2y = 3$ , such as $(-3, 0)$ and $(1, 2)$ , because both equations describe the same line.

Now you will find and fix errors in worked solutions. Check the answer in both original equations, and make sure the work found both variables and used both equations. If a check fails, read the work line by line and find the first step that does not keep the solutions. Watch for these error types:

Multiplying only one side, or only some terms, of an equation.
Subtracting the equations when adding was needed, or adding when subtracting was needed.
Stopping after finding only one variable. A solution is an ordered pair, so substitute that value into an original equation to find the other variable, then check both equations.
Substituting an expression back into the same equation it came from, then using the true statement it gives to count solutions.

Worked example

Example 4: Find and correct an error

A student solved the system $3x + 2y = 20$ and $x - y = 5$ . Line 1: multiply the second equation by $2$ to get $2x - 2y = 5$ . Line 2: add that to the first equation to get $5x = 25$ , so $x = 5$ . Line 3: substitute into $x - y = 5$ to get $5 - y = 5$ , so $y = 0$ . Find and correct the error.

Check the student's answer, (5, 0), in both original equations. It works in x - y = 5 but fails in 3x + 2y = 20, so the work contains an error.
$5 - 0 = 5 \quad \text{but} \quad 3(5) + 2(0) = 15 \ne 20$
Find where it goes wrong: Line 1. The student multiplied the left side by 2 but left the right side as 5.
$2(x - y) = 2(5) \;\Rightarrow\; 2x - 2y = 10, \text{ not } 2x - 2y = 5$
Why it fails: multiplying only one side is not the same operation on both sides, so the solutions change. Dividing both sides of 2x - 2y = 5 by 2 shows that it says x - y = 2.5, a different line from x - y = 5.
$2x - 2y = 5 \;\Rightarrow\; x - y = 2.5$
Correct the work. Add 2x - 2y = 10 to the first equation; 2y and -2y are opposites, so they cancel. Then divide both sides by 5, which is not zero.
$(3x + 2y) + (2x - 2y) = 20 + 10 \;\Rightarrow\; 5x = 30 \;\Rightarrow\; x = 6$
Find y. Substitute x = 6 into x - y = 5, subtract 6 from both sides, then multiply both sides by -1.
$6 - y = 5 \;\Rightarrow\; -y = -1 \;\Rightarrow\; y = 1$
Check (6, 1) in both original equations. Both statements are true.
$3(6) + 2(1) = 18 + 2 = 20 \quad \text{and} \quad 6 - 1 = 5$

Answer

The error is in Line 1: the right side also had to be multiplied by $2$ . The corrected solution is $(6, 1)$ , the only solution of the system.

Here is a second student's work on $2x + 3y = 13$ and $5x - 3y = 1$ , shown in the table below beside the corrected work. The student's answer, $(-4, 7)$ , passes the check in the first equation but fails in the second: $5(-4) - 3(7) = -20 - 21 = -41$ , not $1$ . The error is in Line 1. The $y$ -coefficients, $3$ and $-3$ , are opposites, so the $y$ -terms cancel only when you add the equations. Subtracting is allowed, but it gives $3y - (-3y) = 6y$ , so subtracting correctly leads to $-3x + 6y = 12$ . The student wrote $-3x = 12$ instead, dropping the $6y$ . That equation does not follow from the system: at the true solution, $(2, 3)$ , its left side is $-3(2) = -6$ , not $12$ . So every value after Line 1 is wrong.

The second student's work beside the corrected work

Line	Student's work	Corrected work
1 (combine the equations)	$(2x + 3y) - (5x - 3y) = 13 - 1$ , written as $-3x = 12$	$(2x + 3y) + (5x - 3y) = 13 + 1$ , so $7x = 14$
2 (solve for $x$ )	Divide both sides by $-3$ : $x = -4$	Divide both sides by $7$ : $x = 2$
3 (find $y$ from $2x + 3y = 13$ )	$2(-4) + 3y = 13$ , so $-8 + 3y = 13$ , $3y = 21$ , $y = 7$	$2(2) + 3y = 13$ , so $4 + 3y = 13$ , $3y = 9$ , $y = 3$
4 (check both equations)	$2(-4) + 3(7) = 13$ , but $5(-4) - 3(7) = -41 \ne 1$	$2(2) + 3(3) = 13$ and $5(2) - 3(3) = 1$

To correct the slip, substitute $x = 7 - 2y$ into the other equation, $3x - 2y = 5$ . That gives $3(7 - 2y) - 2y = 5$ , so $21 - 6y - 2y = 5$ and $21 - 8y = 5$ . Subtract $21$ from both sides to get $-8y = -16$ , then divide both sides by $-8$ , which is not zero, to get $y = 2$ . Then $x = 7 - 2(2) = 3$ . Check in both original equations: $3 + 2(2) = 7$ and $3(3) - 2(2) = 9 - 4 = 5$ . The system has exactly one solution, $(3, 2)$ , so its lines cross at exactly one point. Because the $y$ -coefficients, $2$ and $-2$ , are opposites, adding the equations to get $4x = 12$ would have been even quicker.

Takeaway: Use substitution when a variable is alone or has coefficient $1$ or $-1$ , and elimination when a variable's coefficients are equal or opposite or can be made so by multiplying a whole equation by a nonzero number. To count solutions, combine the two different equations, by substituting from one into the other or by adding or subtracting the equations or nonzero multiples of them. Then one value for each variable means exactly one solution (the lines cross once), a true statement such as $0 = 0$ means infinitely many solutions (the same line), and a false statement such as $0 = 5$ means no solution (parallel, distinct lines). Check every answer in both original equations, and when a check fails, find the first step that changed the solutions.