## C2. Equations and Inequalities

### Specific Expectations

#### Variables and Expressions

**expressions:**- 4
*m*+ 3*m*− (+2*m*) − (−4*m*) - −5
*x*+ 4*y*− (−3*x*) + (−*y*) - (4
*x*+ 3*y*) + (5*x*+ 2*y*) - (4
*x*+ 3*y*) + (5*x*− 2*y*) - (−4
*x*+ 3*y*) + (−5*x*+ 2*y*) - (−4
*x*− 3*y*) + (−5*x*− 2*y*)

- 4
**adding binomials with pattern blocks:***h*represents the cost of one yellow hexagon*t*represents the cost of one green triangle*h*+*h*+*h*+*h*+*h*+*h*+*h*= 7*h*→ total cost for the hexagons in 1 tile- 2
*t*+ 2*t*+ 2*t*+ 2*t*+ 2*t*+ 2*t*= 12*t*→ total cost for the green triangles in 1 tile - Total cost of one tile = 7
*h*+ 12*t* - Total cost of two tiles = (7
*h*+ 12*t*) + (7*h*+ 12*t*) = 14*h*+ 24*t*

**adding and subtracting monomials with algebra tiles:**- −5
*x*+ 4*y*− (−3*x*) + (−*y*):

- −5

- A monomial with a degree of 1 has a variable with an exponent of one. For example, the exponent of
*m*for the monomial 2*m*is 1. When the exponent is not shown, it is understood to be one. - A binomial with a degree of 1 consists of two terms (two binomials) in which at least one of the terms has a variable with an exponent of one (e.g., 2
*m*+ 5 or 2*m*+ 5*n*). - Only like terms can be combined when monomials and binomials are added together. For example:
- 5
*m*+ (−3*m*+ 4*n*) - = 5
*m*+ (−3*m*) + 4*n* - = 2
*m*+ 4*n*

- 5
- Monomials with a degree of 1 with the same variables can be subtracted (e.g., −10
*y*− 8*y*= −18*y*). - Monomials can be subtracted in different ways. One way is to compare them and determine the missing addend (e.g., 3
*m*+ ? = 7*m*). Another way is to remove them from the expression representation. - Strategies for performing operations with integers can also be used to add and subtract monomials and binomials.

*Note*

- Examples of monomials with a degree of 2 are
*x*^{2}and*xy*. The reason that*xy*has a degree of 2 is because both*x*and*y*have an exponent of 1. The degree of the monomial is determined by the sum of all the exponents of its variables. - Visual representations can support students’ understanding of combining like terms.
- When adding binomials, brackets are used around the expressions to show that they are binomials.

Arrange students in pairs. Have each student in a pair create a tile using the same type of pattern block. Have each student write an algebraic expression to represent the cost of their tile. Next, have the students in each pair determine the difference between the costs of their tiles. For example, tile A can be represented by 7*h* and tile B can be represented by 14*h*, where *h* represents the cost of each hexagon:

Have students use pattern blocks or algebra tiles to model adding a variety of binomials. It is important for students to see that only like terms can be simplified. For example:

- Add 3
*x*+ 1 and*x*+ 3 using algebra tiles:

- Add (−2
*x*+ 3) and (*x*+ 1) using algebra tiles:

C2.2

evaluate algebraic expressions that involve rational numbers

**algebraic expressions:**- 2
*l*+ 2*w* *bh**bh*÷ 2 or*lwh**πr*^{2}- 2
*πr* *πd*- 3
*m*^{2}*n*– 1

- 2
**evaluating expressions:**- 7
*h*+ 12*t*represents the cost of the tile, where*h*represents the cost of one hexagon and*t*represents the cost of one green triangle:

- 7

if 1*h* = $0.75 and 1*t *= $0.25, then

7*h *+ 12*t *= 7($0.75) + 12($0.25)

= $5.25 + $3.00

= $8.25

- To evaluate an algebraic expression, the variables are replaced with numerical values and calculations are performed based on the order of operations.

*Note*

- When students are working with formulas, they are evaluating expressions.
- Replacing the variables with numerical values often requires the use of brackets. For example, the expression
*m*becomes (*m*) and then () when*m*=. The operation between and () is understood to be multiplication. - Many coding applications involve algebraic expressions being evaluated. This may be carried out in several steps. For example, the instruction: “input ‘the radius of a circle’, radiusA” is instructing the computer to define the variable “radiusA” and store whatever the user inputs into the temporary location called radiusA. The instruction: “input ‘the height of the cylinder’, heightA” is instructing the computer to define the variable “heightA” and store whatever the user inputs into the temporary location called heightA. The instruction: “calculate 3.14*radiusA^2*heightA, volumeA” instructs the computer to take the value that is stored in radiusA and multiply it by itself, then multiply it by the value stored in heightA, and then store that result in the temporary location, which is another variable called “volumeA”. (The caret symbol (^) is used with many forms of technology as the exponent symbol.)

Have students evaluate the algebraic expressions for the tile or picture they created for C2.1, Sample Task 1 when given values for the shapes; for example, the hexagons cost $10.75 each and the trapezoids cost $8.50 each.

Formulas have algebraic expressions. Have students evaluate a variety of formulas, including those introduced in Strand E: Spatial Sense and the Science and Technology curriculum, when appropriate.

#### Equalities and Inequalities

C2.3

solve equations that involve multiple terms, integers, and decimal numbers in various contexts, and verify solutions

**equations:**- 5
*x*− 10 = −3*x*+ 6 - −5
*x*− 3*x*= 10 − (−6) - 3.2
*x*+ 4.1 = 7.3 - 14.5 − 2.5 = 6.5
*x*− 2*x* *x*^{2}= 3^{2}+ 4^{2}

- 5
**solving using algebra tiles and the balance model:**

- Equations are mathematical statements such that the expressions on both sides of the equal sign are equivalent.
- In equations, variables are used to represent unknown quantities.
- There are many strategies to solve equations, including guess-and-check, the balance model, and using the reverse flow chart.
- Equations need to be simplified in order to use the strategy of using a reverse flow chart to solve equations like − 2 = −10. The first diagram shows the flow of operations performed on the variable m to produce the result −10. The second diagram shows the reverse flow chart, or flow of the reverse operations, in order to identify the value of the variable
*m*.

- Formulas are equations in which any of the variables can be solved for. When solving for a variable in a formula, values for the variables are substituted in and then further calculations maybe needed depending on which variable is being solved for. For example, for
*A*=*lw*, if*l*= 10.5, and*w*= 3.5, then*A*= (10.5)(3.5) = 36.75. If*A*= 36.75 and*l*= 10.5, then 36.75 = 10.5*w*, and this will require dividing both sides by 10.5 to solve for*w*.

*Note*

- The flow chart used in coding is different from the reverse flow chart that can be used to solve equations.
- Many coding applications involve formulas and solving equations.

Provide students with equations to solve that require them to collect like terms, such as 5*x *– 10 = –3*x *+ 6.

Once they have simplified an equation, they can use a variety of methods to solve for the unknown value. It is important for them to check their solutions by substituting the value into the equation and verifying that both sides of the equation remain equal. For example, they might use the structure of an LS/RS (left side/right side) check by substituting their solution into both sides of the original equation and then evaluating each side independently. If LS = RS, the solution is correct. If LS ≠ RS, then the solution is incorrect. In the case below, the student has determined that the solution is *x *= 2:

Have students solve equations involving decimal numbers, such as finding the length of one side of a right-angle triangle to the nearest tenth when given the length of the other two sides, or finding side lengths of a shape when given some information about it:

C2.4

solve inequalities that involve integers, and verify and graph the solutions

**sample inequalities and solutions:****Sample 1: 5****x****− 10 < −3****x****+ 6**- Step 1: Isolate the variable
*x*.- Get all the
*x*’s on one side and all the numbers on the other side. - Begin by moving all of the numbers to the right side. This means adding 10 to the left side. To maintain balance, we need to do the same thing on the right side:

- Get all the

5*x *− 10 + 10 < −3*x *+6 + 10

5*x *< −*3x *+ 16

- Now move all of
*x*’s to the left side That means we have to get rid of them on the right side by adding + 3*x*. - To maintain balance, we need to do the same thing on the left side.

5*x* + 3*x* < −3*x* + 3*x* + 16

8*x* < 16

- Step 2: Solve for
*x*.- If
*x*= 1, then 8 × 1 < 16 is true. - If
*x*= 2, then 8 × 2 < 16 is not true. [8 × 2 = 16] - So
*x*must be less than 2: *x*< 2

- If

**Sample 2: 10 − (−6) ≤ −5****x****− 3****x**

10 + 6 ≤ −5*x* – 3*x*

16 ≤ −8*x*

−8*x *≥* *16

−2 ≥ *x*, which can also be expressed as *x* ≤ −2

- An inequality can be solved like an equation, and then values need to be tested to identify those that hold true for the inequality.
- When multiplying or dividing by a negative integer, the inequality sign needs to be reversed in order for the solution to hold true.
- A number line shows the range of values that hold true for an inequality by placing a dot at the greatest or least possible value. An open dot is used when an inequality involves “less than” or “greater than”; if the inequality includes the equal sign (=), then a closed dot is used.

*Note*

- Inequalities that involve multiple terms may need to be simplified before they can be solved.
- The solution for an inequality that has one variable, such as −2
*x*+ 3*x*< 10, can be graphed on a number line.

Have students solve various inequalities that involve simplifying like terms. For example,

- 5
*x*− 10 < −3*x*+ 6 - 10 − (−6) ≤ −5
*x*− 3*x*

A possible strategy for solving inequalities is to first solve the inequality as an equality, then test numbers greater than and less than the solution to the equality in the original inequality to determine the range of numbers for which it holds. Reinforce the proper use of closed or open circles to represent solutions.