No Solutions or Infinitely Many: Build and Check | Grade 8 Math | LevelKite

You already know how to transform a linear equation and read its final line to tell how many solutions it has. Now you will use that skill in two new ways. You will build equations that have no solutions or infinitely many solutions, and you will find and fix mistakes in someone else's work.

In this lesson, the variable $x$ can be any real number, which means any number on the number line. These are the permitted values of $x$ . A solution is a permitted value of $x$ that makes the equation true.

An equation claims that two expressions have the same value. Two expressions are equivalent when they have the same value for every permitted value of the variable.

When the two sides are equivalent, the claim is true for every permitted value. An equation like that is an identity. Here it has infinitely many solutions, because every real number is a solution.
When the two sides never have the same value, the claim is false for every permitted value. An equation like that is a contradiction. It has no solutions.

The table below tests three expressions at four values of $x$ . In the first row, $x = -2$ , so the entries are $2(-2 + 3) = 2(1) = 2$ , then $2(-2) + 6 = -4 + 6 = 2$ , then $2(-2) + 1 = -4 + 1 = -3$ . Look for expressions that match in every row and expressions that match in no row.

Three expressions tested at four values of x

Value of x	2(x + 3)	2x + 6	2x + 1
$-2$	$2$	$2$	$-3$
$0$	$6$	$6$	$1$
$0.5$	$7$	$7$	$2$
$5$	$16$	$16$	$11$

In every row, $2(x + 3)$ and $2x + 6$ have the same value, and $2(x + 3)$ is exactly $5$ more than $2x + 1$ .

Four rows cannot prove what happens for every real number. The distributive property can: $2(x + 3) = 2 \cdot x + 2 \cdot 3 = 2x + 6$ for every real number $x$ . That lets you build two equations:

$2(x + 3) = 2x + 6$ is an identity. Its sides are equivalent, so every real number is a solution.
$2(x + 3) = 2x + 1$ is a contradiction. Its left side is equivalent to $2x + 6$ , which is always $5$ more than $2x + 1$ because $6 - 1 = 5$ . No real number is a solution.

Once the left side is written as $2x + 6$ , both equations have the same $x$ -term, $2x$ , on each side. The number that multiplies the variable is the coefficient, here $2$ . The equations differ only in their constants, the numbers with no variable.

Key concept

Match the x-terms, then let the constants decide

Let $x$ be any real number. Simplify each side of an equation correctly, so each side becomes an equivalent expression: $ax + b$ on the left and $cx + d$ on the right. Here $a$ and $c$ are the coefficients of $x$ , and $b$ and $d$ are the constants.

Infinitely many solutions: choose $a = c$ and $b = d$ . The sides are equivalent, so the equation is an identity and every real number is a solution.
No solutions: choose $a = c$ and $b \ne d$ . Subtracting $ax$ from both sides keeps the solutions the same and leaves $b = d$ , which is false. The equation is a contradiction.

Test a value in the original equation: it must work in an identity and fail in a contradiction. If $a \ne c$ , there is exactly one solution, so the equation is neither kind.

ax + b = cx + d

Worked example

Build an equation with no solutions

Write an equation with no solutions whose left side is $3(x + 4)$ . The variable $x$ can be any real number.

Multiply each term inside the parentheses by 3. This is the distributive property, so the new expression is equivalent to the old one: both have the same value for every real number x.
$3(x + 4) = 3 \cdot x + 3 \cdot 4 = 3x + 12$
For no solutions, the right side needs the same coefficient of x, which is 3, and a different constant. Any constant other than 12 works. Choose 7.
$3x + 7$
Write the equation, keeping the original left side.
$3(x + 4) = 3x + 7$
Confirm the number of solutions. Replace the left side with the equivalent expression 3x + 12, then subtract 3x from both sides. Subtracting the same quantity from both sides keeps the solutions the same.
$3x + 12 - 3x = 3x + 7 - 3x \quad\text{so}\quad 12 = 7$
Read the final statement. It has no x in it, and 12 does not equal 7, so it is false for every real number x. The equation is a contradiction.
Test one value in the original equation. With x = 2, the left side is 18 and the right side is 13, so 2 is not a solution. One failed test cannot prove there are no solutions. Steps 4 and 5 prove that. But a value that worked would reveal a mistake.
$3(2 + 4) = 3(6) = 18, \qquad 3(2) + 7 = 6 + 7 = 13$

Answer

One equation with no solutions is $3(x + 4) = 3x + 7$ . Replacing $7$ with any other constant except $12$ also gives no solutions.

Worked example

Complete an equation with infinitely many solutions

Find the numbers $p$ and $q$ that make $6x - 2(x - 5) = px + q$ have infinitely many solutions. The variable $x$ can be any real number.

Simplify the left side. Multiply each term inside the parentheses by negative 2. Negative 2 times x is negative 2x, and negative 2 times negative 5 is positive 10. The result is equivalent to the original left side for every real number x.
$6x - 2(x - 5) = 6x - 2x + 10$
Combine the two x-terms: 6x minus 2x is 4x.
$6x - 2x + 10 = 4x + 10$
For infinitely many solutions, the right side must be equivalent to the left side, so match the coefficient of x and match the constant. A different coefficient would give exactly one solution, and a different constant would give none.
$p = 4, \qquad q = 10$
Confirm the number of solutions. Subtract 4x from both sides of 4x + 10 = 4x + 10, which keeps the solutions the same. The result, 10 = 10, is true for every real number x, so the equation is an identity.
$4x + 10 - 4x = 4x + 10 - 4x \quad\text{so}\quad 10 = 10$
Test one value in the original equation. With x = 3, both sides equal 22, so 3 is a solution, as it must be when every real number is a solution.
$6(3) - 2(3 - 5) = 18 - 2(-2) = 18 + 4 = 22, \qquad 4(3) + 10 = 12 + 10 = 22$

Answer

With $p = 4$ and $q = 10$ , the equation is $6x - 2(x - 5) = 4x + 10$ . It has infinitely many solutions, because every real number is a solution.

Now use the same ideas to check someone else's work. When the number of solutions someone finds is wrong, look for two kinds of error:

A step error: a line does not follow from the line before it. A step is valid when it replaces an expression with an equivalent one, adds or subtracts the same quantity on both sides, or multiplies or divides both sides by the same nonzero number. You can never divide by zero, and that rules out dividing by $x$ , because $x$ might be $0$ .
A reading error: every step is valid, but the final line is read wrongly, as in the warning above.

A quick test is a good place to start. Always test in the original equation, because a later line may already contain a mistake. For the linear equations in this lesson, a claim of infinitely many solutions means every real number is a solution, so any value you test must work. A claim of no solutions means any value you test must fail. A claim of one solution means that value must work.

One value that breaks the claim proves the claim wrong. A test that agrees with the claim does not prove it, so still check each step and how the final line was read. The table below shows three short cases.

Three wrong conclusions and how to correct them

The work	What went wrong	Corrected conclusion
$x + 4 = x + 9$ becomes $4 = 9$ . Claim: infinitely many solutions.	Reading error. The step is valid, but $4 = 9$ is false for every value of $x$ .	No solutions. For example, $x = 1$ gives $1 + 4 = 5$ but $1 + 9 = 10$ .
$7x = 0$ becomes $x = 0$ . Claim: no solutions.	Reading error. The step is valid, but $0$ is a real number, so $x = 0$ names a solution.	One solution, $x = 0$ . Test it in the original: $7(0) = 0$ is true.
$5x = 2x$ becomes $5 = 2$ after dividing both sides by $x$ . Claim: no solutions.	Step error. $x$ might be $0$ , and you can never divide by zero. In fact $x = 0$ works: $5(0) = 2(0)$ .	Subtract $2x$ from both sides: $3x = 0$ . Divide both sides by $3$ , which is not zero: $x = 0$ . One solution.

Worked example

Find and correct an error

Jordan decided how many solutions $4(x - 3) + 2x = 6x - 3$ has, where $x$ can be any real number. Jordan's work: Line 1: $4x - 3 + 2x = 6x - 3$ . Line 2: $6x - 3 = 6x - 3$ . Line 3: $-3 = -3$ . Conclusion: infinitely many solutions. Find Jordan's error and correct it.

Start with a test in the original equation. If Jordan were right, every real number would be a solution. Try x = 0. The left side is negative 12 and the right side is negative 3, so 0 is not a solution. Jordan's conclusion is wrong.
$4(0 - 3) + 2(0) = 4(-3) + 0 = -12, \qquad 6(0) - 3 = 0 - 3 = -3$
Check each step. Line 1 should replace 4(x - 3) with an equivalent expression, so the 4 must multiply both terms inside the parentheses. Four times x is 4x, and 4 times 3 is 12, which is still subtracted. Jordan multiplied only the x. This is a step error. Lines 2 and 3 follow correctly from Line 1.
$4(x - 3) = 4 \cdot x - 4 \cdot 3 = 4x - 12$
Correct Line 1, then combine the x-terms on the left side: 4x plus 2x is 6x.
$4x - 12 + 2x = 6x - 3 \quad\text{so}\quad 6x - 12 = 6x - 3$
Subtract 6x from both sides, which keeps the solutions the same.
$6x - 12 - 6x = 6x - 3 - 6x \quad\text{so}\quad -12 = -3$
Read the final statement. Negative 12 does not equal negative 3, so the statement is false for every real number x. The equation is a contradiction. This fits step 1, where the test value 0 failed.

Answer

Jordan's error is in Line 1: $4(x - 3)$ is equivalent to $4x - 12$ , not $4x - 3$ . The corrected work ends in the false statement $-12 = -3$ , so the equation has no solutions.

The big idea: For a linear equation in which $x$ can be any real number, simplify both sides correctly and compare them. To build an identity, make the sides equivalent, with the same coefficient of $x$ and the same constant, so every real number is a solution. To build a contradiction, keep the same coefficient of $x$ and use different constants, so no real number is a solution.

To check a claim about the number of solutions, make sure every step is valid: an equivalent expression, or the same operation on both sides, never dividing by zero. Then read the final line for what it truly says, and test the conclusion in the original equation.