Click and drag algebra tiles to move them. You can add tiles from the bottom to the scale or you can move tiles from the scale to the bottom of the applet to take them off the scale.

Click the reset button in the top right corner to go back to the equation 3x+ 1=7.

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What kinds of things can you add (or remove) on one side, without changing the other side, and keep the scale balanced?
Use the scale to find the value of one +x tile. Explain your process. (Remember the scale must stay balanced.)

Recall this scenario:

"You have 2 times the number of tokens as your friend. After giving away 4 tokens, you have 16 tokens left. How many tokens did your friend start with?"

We represented the scenario using pictures like this:

Pictorial model representing 2x-4=16: 2 pouches, then 4 greyed-out tokens representing giving them away, then = 16 tokens.

We can use the pictures to help us determine how many tokens your friend started with.

Pictorial model representing 2x-4+4=16+4: 2 pouches, then 4 greyed-out tokens representing giving them away, then = 16 tokens. 4 tokens is then added to both sides.

We can start by adding tokens to both sides of the equation to create zero pairs on the left side.

Two pictorial models. The first represents 2x=20, showing 2 pouches = 20 tokens. The second pictorial model represents x=10, where 1 pouch = 10 tokens is shown.

Once the zero pairs are removed and we combine the tokens on the right we are left with 2 bags on the left side and 20 tokens on the right side.

To determine our final answer, we can divide the tokens on the right side into 2 equal groups to find that one bag holds 10 tokens.

Your friend started with 10 tokens.

We can also use algebra tiles to represent the scenario. The algebra tiles would look like this:

Algebra tiles representing 2x-4=16: two positive variable tiles and four negative unit tiles is equal to sixteen positive unit tiles.

We can start by adding +1 tiles to both sides of the equation to create zero pairs on the left side.

Algebra tiles representing 2x-4=16: two positive variable tiles and four negative unit tiles is equal to sixteen positive unit tiles. Four positive unit tiles are added to both sides.

Two sets of algebra tiles. The first one represents 2x=20, showing 2 positive variable tiles is equal to 20 positive unit tiles. The second one represents x=10, showing 1 positive variable tile is equal to 10 positive unit tiles.

Once we remove the zero pairs, we are left with two +x tiles on the left side and twenty+1 on the right side.

To solve, we can divide the +1 on the right side into 2 equal groups.

Again we find that your friend started with 10 tokens.

Examples

Example 1

Consider the following algebra tiles:

A set of tiles representing the equation 2x-3=5. 2 positive variable tiles and 3 negative unit tiles on the right side, and 5 positive unit tiles on the right.

What should we add to the left side and right side of the equation to keep only the x tiles on the left side of the equation?

Worked Solution

Create a strategy

Keep the two sides balanced by adding tiles to have zero pairs and cancel the unit tiles on the left side.

Apply the idea

Add 3 positive unit tiles to both sides of the equation to cancel out the 3 negative unit tiles.

A set of tiles representing the equation 2x-3+3=5+3: On the left side are 2 positive variable tiles, 3 negative unit tiles and 3 positive unit tiles; on the right are 5 positive unit tiles and an additional 3 positive unit tiles.

Draw the final number of algebraic tiles and write the equation to solve for x.

Worked Solution

Create a strategy

Count the final number of tiles after canceling out the unit tiles.

Apply the idea

A set of tiles representing the equation 2x=8: On the left are 2 positive variable tiles and on the right are 8 positive unit tiles.

Find the value of x.

Worked Solution

Create a strategy

We divide the tiles into equal groups on both sides of the equation to find the equivalent of a single x tile.

Apply the idea

Divide by 2 groups.

A set of tiles representing the equation x=4: On the left is 1 positive variable tile and on the right are 4 positive unit tiles.

Example 2

You and your friends shared 2 pizzas. You also got a pitcher of soda for \$6.75. You paid a total of \$31.75.

Create a pictorial model to represent this scenario.

Worked Solution

Create a strategy

The unknown value is the cost of pizza.

Apply the idea

We can use a pictorial model like this:

Pictorial model: Two pizzas + 5 USD 1 USD + three 0.25 USD bills/coins = six 5 USD 1 USD three 0.25 USD.

Find the cost of each pizza using your model.

Worked Solution

Create a strategy

We can represent the scenario using the equation 2p+6.75=\$31.75 and use the pictorial model to solve it.

Apply the idea

Remove \$6.75 of the money from both sides of the equation.

Pictorial model: Two pizzas + 5 USD, 1 USD and three 0.25 USD bills/coins = six 5 USD, 1 USD and three 0.25 USD. 6.75USD is greyed-out from both sides to represent taking them out.

Now we know 2 pizzas cost \$25.

Organize the remaining money into two equal sized groups. We will need to break one of the \$5 bills into four \$1 bills and four quarters to divide the money up evenly.

Pictorial model: 2 pizzas = two groups of 12.50 USD

Remove one pizza and one group of money from each side.

We can now see that one pizza costs \$12.50.

Idea summary

Algebra tiles and pictorial allow us to represent an equation visually. When using these models, we can look for zero pairs to help isolate the variable and solve the equation.

It is important to ensure that we are keeping the two sides of the equation balanced, so what we do to one side, we must do to the other.

Solve equations using properties

Algebra tiles and pictorial models are not always a realistic or efficient way to solve an equation. We can use the properties of equality to solve an equation. These properties require us to apply the same operation on both sides of the equation as we try to isolate the variable. This keeps the equation balanced.

The properties of equality used to solve equations include:

Addition property of equality	\text{If } a=b, \text{then } a+c=b+c
Subtraction property of equality	\text{If } a=b, \text{then } a-c=b-c
Multiplication property of equality	\text{If } a=b, \text{then } ac=bc
Division property of equality	\text{If } a=b \text{ and } c \neq0, \text{then } \dfrac{a}{c}=\dfrac{b}{c}
Substitution property of equality	\text{If } {a=b}, \text{ then } b \text{ may be substituted for } a \text{ in any expression}

The properties of real numbers are also useful in solving equations. The ones we use most often are:

Commutative property of addition	a+b=b+a
Commutative property of multiplication	a \cdot b=b \cdot a
Associative property of addition	a+\left(b+c\right) =\left(a+b\right)+c
Associative property of multiplication	a\cdot \left(b\cdot c\right)=\left(a\cdot b\right)\cdot c
Additive identity	\text{If } {a=b}, \text{ then } a+0=b \text{ and } a=b+0
Multiplicative identity	\text{If } {a=b}, \text{ then } a\cdot 1=b \text{ and } a=b \cdot 1
Additive inverse	a+\left(-a\right)=0
Multiplicative inverse	a \cdot \dfrac{1}{a}=1

Examples

Example 3

Solve the following equation: 8m+9=65

Worked Solution

Create a strategy

Use the properties of equality to isolate the variable.

Apply the idea

\displaystyle 8m+9	\displaystyle =	\displaystyle 65	Start with the given equation
\displaystyle 8m+9-9	\displaystyle =	\displaystyle 65-9	Subtraction property of equality
\displaystyle 8m	\displaystyle =	\displaystyle 56	Evaluate the subtraction
\displaystyle \frac{8m}{8}	\displaystyle =	\displaystyle \frac{56}{8}	Division property of equality
\displaystyle m	\displaystyle =	\displaystyle 7	Evaluate the division

Reflect and check

We can check our answer by substituting the value we found for m, back into the equation.

\displaystyle 8\left(7\right)+9	\displaystyle =	\displaystyle 65	Substitution property
\displaystyle 56+9	\displaystyle =	\displaystyle 65	Evaluate the multiplication
\displaystyle 65	\displaystyle =	\displaystyle 65	Evaluate the addition

When we substitute the value of m into the equation, we are left with 65=65. This is true, so our solution is correct.

Example 4

Solve the following equation: \dfrac{x-4.2}{3.1} = 8.2. Round to 2 decimal places if necessary.

Worked Solution

Create a strategy

Use the properties of equality to isolate the variable.

Apply the idea

\displaystyle \dfrac{x-4.2}{3.1}	\displaystyle =	\displaystyle 8.2	Start with the given equation
\displaystyle \dfrac{x-4.2}{3.1} \cdot 3.1	\displaystyle =	\displaystyle 8.2 \cdot 3.1	Multiplication property of equality
\displaystyle x-4.2	\displaystyle =	\displaystyle 8.2 \cdot 3.1	Multiplicative inverse
\displaystyle x-4.2	\displaystyle =	\displaystyle 25.42	Evaluate the multiplication
\displaystyle x-4.2+4.2	\displaystyle =	\displaystyle 25.42 + 4.2	Addition property of equality
\displaystyle x	\displaystyle =	\displaystyle 29.62	Evaluate the addition

Reflect and check

We can check our answer by substituting the value we found for x back into the equation.

\displaystyle \dfrac{\left(29.62\right)-4.2}{3.1}	\displaystyle =	\displaystyle 8.2	Substitution property
\displaystyle \dfrac{25.42}{3.1}	\displaystyle =	\displaystyle 8.2	Evaluate the subtraction
\displaystyle 8.2	\displaystyle =	\displaystyle 8.2	Evaluate the division

When we substitute the value of x into the equation, we are left with 8.2=8.2. This is true, so our solution is correct.

Example 5

Determine whether the given value is a solution to the following equations.

5x+3=23 where x=4

Worked Solution

Create a strategy

To determine if the value is a solution, we will use the substitution property to substitute 4 into the equation for x.

Apply the idea

\displaystyle 5\left(4\right)+3	\displaystyle =	\displaystyle 23	Substitution Property
\displaystyle 20+3	\displaystyle =	\displaystyle 23	Evaluate the multiplication
\displaystyle 23	\displaystyle =	\displaystyle 23	Evaluate the addition

Because 23=23, x=4 is a solution to the equation 5x+3=23.

-\dfrac{5}{2}f+ \dfrac{1}{2}= \dfrac{11}{2} where f=-6

Worked Solution

Create a strategy

To determine if the value is a solution, we will use the substitution property to substitute -6 in the equation for f.

Apply the idea

\displaystyle -\dfrac{5}{2}(-6)+ \dfrac{1}{2}	\displaystyle =	\displaystyle \dfrac{11}{2}	Substitution Property
\displaystyle 15+\dfrac{1}{2}	\displaystyle =	\displaystyle \dfrac{11}{2}	Evaluate the multiplication
\displaystyle \dfrac{31}{2}	\displaystyle =	\displaystyle \dfrac{11}{2}	Evaluate the addition

Because \dfrac{31}{2} \neq \dfrac{11}{2}, f=-6 is not a solution to the equation -\dfrac{5}{2}f + \dfrac{1}{2}=\dfrac{11}{2}.

Idea summary

We can use the properties of real numbers and properties of equality to solve two-step equations.

When solving two-step equations:

We want to get the variable by itself on one side of the equals sign
To keep everything balanced, we must do the same operations to both sides by applying the properties of equality

Outcomes

7.PFA.3

The student will write and solve two-step linear equations in one variable, including problems in context, that require the solution of a two-step linear equation in one variable.

7.PFA.3a

Represent and solve two-step linear equations in one variable using a variety of concrete materials and pictorial representations.

7.PFA.3b

Apply properties of real numbers and properties of equality to solve two-step linear equations in one variable. Coefficients and numeric terms will be rational.

7.PFA.3c

Confirm algebraic solutions to linear equations in one variable.

4.02 Solve two-step equations

Ideas

Solve equations using representations

Exploration

Examples

Example 1

Example 2

Solve equations using properties

Examples

Example 3

Example 4

Example 5

Outcomes

7.PFA.3

7.PFA.3a

7.PFA.3b

7.PFA.3c

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