M6|L22 Combining Like Terms (Part 3)Sign out

Question

Write the expression in simplest form.
Use the distributive property to write an equivalent expression without parentheses.
(2y+12z)−(7y+2z)
= 2y + 12z +
+
Great job!
Next, write an equivalent expression by grouping all of the like variables together.
=
2y
+
−7y
+
12z
+
−2z
Good work!
Now rewrite the expression with the fewest possible terms.
(2y+12z)−(7y+2z)
= 2y + 12z + −7y + −2z
= 2y + −7y + 12z + −2z
=
Excellent!
Factor to rewrite the expression another way. To start factoring, choose the greatest common factor of −5y and 10z.
(2y+12z)−(7y+2z)
= 2y + 12z + −7y + −2z
= 2y + −7y + 12z + −2z
= −5y+10z

6

5

4
Good work!
Now find the factors that, when multiplied by 5, equal −5y and 10z and fill them in the parentheses to write the factored expression.
(2y+12z)−(7y+2z)
= 2y + 12z + −7y + −2z
= 2y + −7y + 12z + −2z
= −5y+10z
= 5(
+
)
Excellent!
Rewrite 5(−y+2z) using subtraction.
(2y+12z)−(7y+2z)
= 2y + 12z + −7y + −2z
= 2y + −7y + 12z + −2z
= −5y+10z
= 5(−y+2z)
= 5(
−
)

Answer 1

To rewrite \(5(−y + 2z)\) using subtraction, you can express it as follows:

\[ 5(-y + 2z) = 5(-y) + 5(2z) = -5y + 10z \]

To rephrase \(5(-y + 2z)\) using subtraction in the parentheses, you could write it as:

\[ 5(-y + 2z) = 5(-y + 2z) = 5(-y + 2z) \]

So this can also be rewritten to show the subtraction clearly:

\[ 5(2z - y) \]

Putting it all together with the context you initially provided:

Starting from:

\[ (2y + 12z) - (7y + 2z) \]

You have simplified it to:

\[ = 5(-y + 2z) \]

And then rewritten with subtraction as:

\[ = 5(2z - y) \]

Both forms \(5(-y + 2z)\) and \(5(2z - y)\) are valid representations of the same expression.