# Classifying Solutions of Linear Equations04:16 minutes

Video Transcript

## TranscriptClassifying Solutions of Linear Equations

Bert Cutler and the high school drama club are having a Murder Mystery party after school in the auditorium. Bert loves solving mysteries, so of course he is playing the detective. Everybody is acting a little suspicious. The butler looks shady, the groundskeeper is definitely hiding something, and the rich old lady is nervously clutching her pearls! So...who-dunnit? Bert goes with his gut and accuses the BUTLER! The players show their cards, but what’s this? Not only is the butler not guilty, but no one else is guilty, either. Somebody really messed up here. There’s no solution to the mystery! The drama club players reshuffle the cards and deal again. Bert accuses the groundskeeper and this time. He's right! The groundskeeper is guilty, but all the other players are guilty too! So, what 's going on? To figure out the mystery inside this mystery, we're going to have to investigate classifying solutions to linear equations. Just like the students' murder mystery game, a linear equation can have NO solution. Let’s look at an example. To calculate the solution to a linear equation, we need to follow the proper steps. First, to get rid of the parentheses, we’ll use the Distributive Property. Then, use opposite operations to isolate the 'x'. Hmm...this answer doesn't look right. 2 is NOT equal to 0. This is a false statement, which means there’s no solution to this linear equation. So, if after following the steps to solve a linear equation, you end up with an answer that looks like something like this, and 'a' and 'b' are not the same number. That means the linear equation has no solution. But there are also cases when a linear equation has an infinite number of solutions. Just like in the game when EVERYONE was guilty, in these cases ANY number can be a solution. Let’s take a look at an example. As always, follow the proper steps to solve the linear equation. First, use the Distributive Property to eliminate the parentheses and then combine the like terms. Look! Both sides of the equal sign are the same! We could simplify that even further by subtracting 2 from both sides, then 5x from both sides, and we end up with an equation that is obviously always going to be true. This means this linear equation has infinitely many solutions. So, if after following the proper steps, you get something that looks like this, or both sides of the equal sign have the same expression. It means there are infinite solutions to the linear equation. The last kind of solution is one that you've seen lots of times: a linear equation with just one solution. By now, you're a pro at this. Use the Distributive Property to write the equation without the parentheses. Combine the like terms and use opposite operations to isolate the variable. As you can see, there's just one solution to this linear equation. To review, there are three different kinds of solutions to linear equations. A linear equation can have no solution. You can recognize this when you end up with an equation like 2 equals 0, which is NOT (never) true. A linear equation can also have an infinite number of solutions. This is when you end up with an equation that is ALWAYS true, like zero equals zero. Finally, there can also be just one solution to a linear equation. These are equations like 5 equals 'x', where a variable is equal to a number. Bert's ready to solve the mystery. This time, he's sure he can identify the guilty person because there should be just one solution to this mystery. Bert takes center stage and accuses...himself! I knew there was something suspicious about that guy!

## Classifying Solutions of Linear Equations Exercise

### Would you like to practice what you’ve just learned? Practice problems for this video Classifying Solutions of Linear Equations help you practice and recap your knowledge.

• #### Explain why the equations have one, infinitely many, or no solutions.

##### Hints

Each linear equation leads to one of the following cases:

• a false equation, like $3=0$. In this case, the equation has no solution.
• an equation which is always true, like $3=3$, independent of the value of the variable. In this case, the equation has infinitely many solutions.
• an equation which is solvable, like $2x=4$. In this case, the equation has only one solution.

Here you see the distributive property.

Here you see an example for solving an equation using opposite operations.

##### Solution

To solve a linear equation you isolate the variable by following the following steps:

• Use the distributive property to multiply parentheses.
• Combine the like terms.
• Use opposite operations to isolate the variable.
Let's try it:

$\begin{array}{rclll} 2(3x+1) & = & ~6x&|&\text{ distributive property}\\ 6x+2&=&~6x\\ \color{#669900}{-6x} & &\color{#669900}{-6x}\\ 2&=&~0 \end{array}$

What's that? This isn't true at all. Thus the equation at the beginning can't be solvable; i.e. the equation has no solution.

Let's have a look at another example:

$\begin{array}{rclll} 3x+2(x+1)&=&~6x+2-x&|&\text{ distributive property}\\ 3x+2x+2&=&~6x+2-x&|&\text{ combine the like terms}\\ 5x+2&=&~5x+2\\ \color{#669900}{-2} & &\color{#669900}{-2}\\ 5x & = & ~5x\\ \color{#669900}{-5x} & & \color{#669900}{-5x}\\ 0 & = & ~0 \end{array}$

This equation is always true. So we can conclude that the starting equation has infinitely many solutions.

$\begin{array}{rclll} 4(5+x)&=&~5(2x+3)-5x&|&\text{ distributive property}\\ 20+4x&=&~10x+15-5x&|&\text{ combine the like terms}\\ 20+4x&=&~5x+15\\ \color{#669900}{-15} & &\color{#669900}{-15}\\ 5+4x & = & ~5x\\ \color{#669900}{-4x} & & \color{#669900}{-4x}\\ 5 & = & ~x \end{array}$

This is a clear solution. Such an equation has only one solution.

• #### Describe how to determine if an equation has one, infinitely many, or no solutions.

##### Hints

This equation has no solution.

This equation has infinitely many solutions.

This equation has just one clear solution; it's $x=1$.

##### Solution

There are only three possibilities for the solutions of a linear equation:

• No solution
• Infinitely many solutions
• One solution
You can recognize the solutions to a linear equation by isolating the variable:

No solution: If you end up with an equation like $a=b$ where $a$ and $b$ are not the same numbers, then you can conclude that the equation has no solution.

Infinitely many solutions: If you end up with an equation like $a=a$, then you can conclude that the equation has infinitely many solutions.

One solution: There exists only one clear solution, then you can conclude that the equation has on solution.

• #### Find equations with one, infinitely many, or no solutions.

##### Hints

Isolate the variable using opposite operations.

Each equation similar to $a=a$ indicates infinitely many solutions.

An equation $a=b$, where $a$ and $b$ are not the same number, indicates no solution.

##### Solution

If you are able to solve an equation uniquely, like $3x=9$, then there is only one solution. In the case of $3x=9$, you have to divide both sides by $3$ to get $x=3$.

Another equation with only one solution is $2(x+1)=4x-(x+2)$:

• Using the distributive property leads to $2x+2=4x-x-2$.
• Next we combine like terms: $2x+2=3x-2$.
• Subtracting $2x$ and $2$ results in $-4=x$.
If you divide $3x$ on both sides of the equation $3x=3x+9$ you get $0=9$. This equation is false. Thus the initial equation has no solutions.

Another equation with no solution is $2(x+1)=3x-(x+2)$:

• With the distributive property we get $2x+2=3x-x-2$.
• Next we combine the like terms: $2x+2=2x-2$.
• Subtracting $2x$ leads to $2=-2$ a false equation.
The two equations left have infinitely many solutions:

Combing like terms in $3x+6x=9x$ leads to $9x=9x$ which is always true.

For $2(x-1)=3x-(x+2)$:

• Using the distributive property and combing like terms, we get $2x-2=2x-2$.
• Subtracting $2x$ and adding $2$ gives the equation $0=0$.
• #### Determine which equations have one, infinitely many, or no solutions.

##### Hints

Simplify each equation using the distributive property and combine like terms.

$x=0$ can be a true a solution to an equation.

There are three equations with one solution.

##### Solution

• $3x+2=2$: subtracting $2$ and dividing by $3$ after leads to $x=0$.
• $3x+2=2x+3$: subtracting $2x$ and $2$ leads to $x=1$.
• $4(x+2)=2(x+4)$ is equivalent to $4x+8=2x+8$. Now subtract $2x$ and $8$ to get $2x=0$. Dividing by $2$ leads to $x=0$.
There are two equations with no solution:

• $3+2x=2x$: subtracting $2x$ leads to $3=0$, which isn't true at all.
• $4(x+2)=2(2x+4)-2$ is equivalent to $4x+8=4x+8-2$. Subtracting $4x$ and combining like terms leads to $8=6$, which is a false equation.
The remaining equations have infinitely many solutions:

• $3x+2x=2x+3x$, which is equivalent to $5x=5x$. Subtract $5x$ to get $0=0$. This equation is always true.
• $4(x+2)=2(2x+4)$ is equivalent to $4x+8=4x+8$. This equation is always true.
• #### Solve the equations which have one solution.

##### Hints

Use the distributive property.

Here you see an example of combining like terms.

Here you see an example of using opposite operations.

##### Solution

To solve linear equations you proceed as follows, perhaps in a different order:

• Use the distributive property.
• Combine like terms.
• Use opposite operations.
Let's have a look at a few examples:

$~$

$\begin{array}{rclll} 3x+4&=&~13\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ 3x & = & ~9\\ \color{#669900}{\div 3} & & \color{#669900}{\div 3}\\ x & = & ~3 \end{array}$

$~$

$\begin{array}{rclll} 3(x+4)&=&~15&|&\text{ distributive property}\\ 3x+12&=&~15\\ \color{#669900}{-12} & &\color{#669900}{-12}\\ 3x & = & ~3\\ \color{#669900}{\div 3} & & \color{#669900}{\div 3}\\ x & = & ~1 \end{array}$

$~$

$\begin{array}{rclll} 3(x+4)-2x&=&~12&|&\text{ distributive property}\\ 3x+12-2x&=&~12&|&\text{ combine the like terms}\\ x+12&=&~12\\ \color{#669900}{-12} & &\color{#669900}{-12}\\ x & = & ~0 \end{array}$

$~$

$\begin{array}{rclll} 3(x+4)+2x&=&~4(x+1)&|&\text{ distributive property}\\ 3x+12+2x&=&~4x+4&|&\text{ combine the like terms}\\ 5x+12&=&~4x+4\\ \color{#669900}{-4x} & &\color{#669900}{-4x}\\ x+12 & = & ~4\\ \color{#669900}{-12} & & \color{#669900}{-12}\\ x & = & ~-8 \end{array}$

• #### Identify which equation describes the story and how many solutions it has.

##### Hints

$2$ times the sum of a number and $3$ is given by $2(x+3)$.

To isolate a variable, use the distributive property, pictured beside, and combine like terms.

##### Solution

For the first story,

• Three times the sum of the unknown number and $2$ is given by $3(x+2)$.
• Three times the number added to $2$ can be written as $3x+2$.
So we get the equation $3(x+2)=3x+2$ which is equivalent to $3x+6=3x+2$. Subtracting $3x$ leads to $6=2$. This is a false equation. Thus we can conclude that the starting equation has no solution.

$~$

For the second story,

• ... twice as old as the sum of his sister Anne's age and $2$ indicates $2(x+2)$.
• Together indicates addition: $x+2(x+2)=16$.
To solve:

$\begin{array}{rclll} x+2(x+2)&=&~16&|&\text{ distributive property}\\ x+2x+4&=&~16&|&\text{ combine the like terms}\\ 3x+4&=&~16\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ 3x & = & ~12\\ \color{#669900}{\div 3} & & \color{#669900}{\div 3}\\ x & = & ~4 \end{array}$

And so this equation has one solution.

$~$

For the last story,

We establish the equation step by step:

• The sum of the number and $4$: $x+4$.
• $7$ times this sum $7(x+4)$.
• Two times the sum of the number and two: $2(x+2)$
• Subtract ... $7(x+4)-2(x+2)$
Let's have a look at the other side of the equation:

• Triple of the number: $3x$.
• The double of the sum of the number and $12$: $2(x+12)$
• Added together we get $3x+2(x+12)$
Now the equation is finished: $7(x+4)-2(x+2)=3x+2(x+12)$.

Using the distributive property and combing like terms gives us

$7x+28-2x-4=3x+2x+24$, and thus $5x+24=5x+24$,

an equation which is true for all $x$; this equation has infinitely many solutions.