Check Your Readiness

1.

Which expression is equivalent to $6(2y-5)$ ?

A.

$12y-5$

B.

$5-12y$

C.

$12y-30$

D.

$30-12y$

Answer:

$12y-30$

Teaching Notes

The distributive property will prove to be an important tool in solving linear equations.

If most students struggle with this item, plan to revisit it before Activity 3 to review using the distributive property. Use hanger diagrams as a context for reviewing the distributive property. Monitor student strategies during the Activity for students who need more practice writing equivalent expressions using the distributive property. Another opportunity to practice using the distributive property appears in Lesson 7.

2.

Which of these expressions is equivalent to $\text-4(5-2x)$ ?

A.

$\text-20+8x$

B.

$\text-20-8x$

C.

$\text-20x+8$

D.

$20x+8$

Answer:

$\text-20+8x$

Teaching Notes

Like the previous question, this question assesses the distributive property. This time, the work involves more difficult multiplication by negative numbers.

If most students struggle with this item, plan to revisit it with students before Activity 2, or use this item as the Warm-up. Ask students to choose a few values to substitute for $x$ to verify their choice of equivalent expressions. (Note: This method is not sufficient for judging equivalent expressions, but will give students practice with operations with rational numbers.)

3.

For each expression, combine like terms, and write an equivalent expression with fewer terms.

$\text-2x+4-5x$
$16+x+5$
$3x-5x+7-2$
$1-7x-3x-1$
$6x+4x$

Answer:

$\text-7x+4$ (or equivalent)
$x+21$ (or equivalent)
$\text-2x+5$ (or equivalent)
$\text-10x$
$10x$

(Equivalent expressions should also be accepted. For example, $21+x$ instead of $x+21$ .)

Teaching Notes

Parts B and E of this problem simply involve recognizing which are “like terms.” Parts A, C, and D require understanding which terms the negative signs apply to. Students will need to be comfortable combining like terms when solving linear equations.

If most students struggle with this item, plan to ask students to write expressions in Activity 3 in different ways, using the hanger diagram to emphasize that $x+4x$ is the same as $5x$ . Each hanger diagram in the lessons offers an opportunity to write equivalent expressions that support students in combining like terms.

4.

For each equation, find a value for $x$ that makes the equation true.

$65=\text-5x$
$10=3x-6$
$3(10-x)=6$
$\frac{5}{4}x=\frac{9}{4}$
$\frac{1}{2}(x-6)=7$
$1.5x-2.4=0.6$

Answer:

$x=\text-13$
$x=\frac{16}{3}$ (or equivalent)
$x=8$
$x=\frac{9}{5}$ (or equivalent)
$x=20$
$x=2$

Teaching Notes

Parts A and E require either distributing the left side or dividing each side by the coefficient as the first step. Part F requires attention to detail with both the decimals and the negative numbers. This is also an opportunity to check in with students’ calculations on decimals and fractions. Encourage students to work on these problems without a calculator.

If most students struggle with this item, plan to revisit this item or one- and two-step equations before this lesson and after students have worked with the hanger diagrams. Work with the hangers should support students as they make decisions about equation-solving moves.

5.

Why is -3 a solution to each of these questions?

$\frac{1}{3}x-6=\text-7$

$2(x+12)=18$

Answer:

Sample response:

$\frac{1}{3}\boldcdot \text-3-6=\text-7$ , and $\text-1 - 6 = \text-7$
$2(\text-3+12)=18$ , $2(9)=18$

Teaching Notes

Before beginning work with systems of equations, students must have a solid understanding of what it means for a value to be a solution to an equation. For each of these questions, it is most efficient to substitute -3 for $x$ and see if the result is a true equation. Some students may instead take algebraic steps to solve each equation. Make sure that students who do the latter are exposed to the substitution strategy, whether through class discussion or individual conversation.

If most students struggle with this item, plan to have students check their solutions by substituting values back into equations. Have them do that beginning in this lesson, and emphasize that a solution to an equation is a value for the variable that makes the equation true.