Worksheet featuring 10 algebra problems where students simplify exponential expressions and match them to answers to color a heart.

Let’s solve each expression in Column 1 and match it to the correct simplified form in Column 2.

We’ll go one by one, simplify carefully, then find the match.

---

Problem 1: $2x^5y \cdot 3x^2y$

Multiply coefficients: $2 \cdot 3 = 6$

Add exponents for same bases:

- $x^5 \cdot x^2 = x^{7}$
- $y \cdot y = y^2$

→ Answer: $6x^7y^2$

Wait — let’s check Column 2. None of them look like that? Let me double-check later. Maybe I made a mistake or need to re-simplify others first.

Actually, let’s hold off on matching until we simplify all of Column 1 and Column 2 expressions. But since the directions say “simplify each expression” and “identify matching answers”, we should simplify both columns and pair them.

But note: The problem says “Simplify each expression. Identify matching answers between Column 1 and Column 2”. So likely, each item in Column 1 simplifies to one item in Column 2 (after simplifying that too).

So let’s simplify ALL expressions in both columns.

---

Simplify Column 1:

1. $2x^5y \cdot 3x^2y$

= $(2\cdot3) \cdot x^{5+2} \cdot y^{1+1} = 6x^7y^2$

2. $(4x^4y)^2 \cdot 2x^3y^4$

First, square the term: $(4x^4y)^2 = 16x^8y^2$

Then multiply: $16x^8y^2 \cdot 2x^3y^4 = 32x^{11}y^6$

3. $\frac{36x^9y^4}{4x^7y^3}$

Coefficients: $36/4 = 9$

Exponents: $x^{9-7} = x^2$, $y^{4-3} = y^1$

→ $9x^2y$

4. $\frac{(2xy^5)^3}{2x^3y^8}$

Numerator: $(2xy^5)^3 = 8x^3y^{15}$

Denominator: $2x^3y^8$

Divide: $8/2 = 4$, $x^{3-3}=x^0=1$, $y^{15-8}=y^7$

→ $4y^7$

5. $(-5x^6y^2)^2 - 12x^{12}y^4$

First term: $(-5)^2 = 25$, $(x^6)^2 = x^{12}$, $(y^2)^2 = y^4$ → $25x^{12}y^4$

Subtract: $25x^{12}y^4 - 12x^{12}y^4 = 13x^{12}y^4$

6. $\frac{6x^{10}y^4}{3x^8y^7}$

Coefficients: $6/3 = 2$

Exponents: $x^{10-8}=x^2$, $y^{4-7}=y^{-3}$

→ $2x^2y^{-3}$ or $\frac{2x^2}{y^3}$

7. $6x^{-1}y^5 \cdot 4x^{-4}y^{-2}$

Coefficients: $6*4=24$

Exponents: $x^{-1 + (-4)} = x^{-5}$, $y^{5 + (-2)} = y^3$

→ $24x^{-5}y^3$ or $\frac{24y^3}{x^5}$

8. $(3x^{-6}y^2)^3 \cdot 2x^{10}y^{-7}$

First part: $3^3 = 27$, $x^{-18}$, $y^6$

Multiply by second part: $27 * 2 = 54$, $x^{-18+10}=x^{-8}$, $y^{6 + (-7)}=y^{-1}$

→ $54x^{-8}y^{-1}$ or $\frac{54}{x^8y}$

9. $\frac{(-2xy)^2 \cdot 10x^3y^{11}}{8x^{10}y^4}$

Numerator: $(-2xy)^2 = 4x^2y^2$, times $10x^3y^{11} = 40x^{5}y^{13}$

Denominator: $8x^{10}y^4$

Divide: $40/8 = 5$, $x^{5-10}=x^{-5}$, $y^{13-4}=y^9$

→ $5x^{-5}y^9$ or $\frac{5y^9}{x^5}$

10. $\frac{8x^3 \cdot 12xy^7}{3x^2y^4} - 15x^2y^3$

First, numerator: $8*12 = 96$, $x^3*x = x^4$, $y^7$

So fraction: $\frac{96x^4y^7}{3x^2y^4} = 32x^{2}y^{3}$

Now subtract: $32x^2y^3 - 15x^2y^3 = 17x^2y^3$

Wait — but looking at Column 2, none have 17? Hmm. Did I misread?

Check original Problem 10:
$\frac{8x^3 \cdot 12xy^7}{3x^2y^4} - 15x^2y^3$

Yes, 8*12=96, divided by 3 is 32. Exponents: x^(3+1-2)=x^2, y^(7-4)=y^3 → 32x²y³ minus 15x²y³ = 17x²y³.

But Column 2 doesn’t have 17... maybe I made an error? Or perhaps the expression is different?

Wait — let me check again: Is it possible the subtraction is outside? Yes, as written.

But let’s look at Column 2 now — maybe some of these simplify to match.

---

Now simplify Column 2 expressions:

Red: $x^4y^8 \cdot 5x^{-9}y$

= $5 \cdot x^{4-9} \cdot y^{8+1} = 5x^{-5}y^9$ → matches our #9!

Orange: $\frac{(6x^2y^3)^2}{4x^2y^5}$

Numerator: $36x^4y^6$

Denominator: $4x^2y^5$

→ $9x^{2}y^{1}$ → matches our #3! ($9x^2y$)

Yellow: $(3x^{-3}y)^3 \cdot 2xy^{-4}$

First: $27x^{-9}y^3 \cdot 2xy^{-4} = 54x^{-8}y^{-1}$ → matches our #8!

Light Green: $\frac{(3x^5y^5)^3}{3x^3y^{11}} + 4x^{12}y^4$

First term: $(27x^{15}y^{15}) / (3x^3y^{11}) = 9x^{12}y^4$

Plus $4x^{12}y^4$ → total $13x^{12}y^4$ → matches our #5!

Dark Green: $\frac{34x^{10}y^9}{2x^8y^6}$

= $17x^{2}y^{3}$ → matches our #10! (I had 17x²y³ — yes!)

Earlier I thought 17 wasn't there, but Dark Green is exactly that.

Light Blue: $\frac{42x^9y^5}{7x^2y^3}$

= $6x^{7}y^{2}$ → matches our #1!

Dark Blue: $3x^{-4}y^9 \cdot 8x^{-1}y^{-6}$

= $24x^{-5}y^{3}$ → matches our #7!

Purple: $(2x^{-2}y^3)^4 \cdot 2x^{19}y^{-6}$

First: $16x^{-8}y^{12} \cdot 2x^{19}y^{-6} = 32x^{11}y^{6}$ → matches our #2!

Pink: $\frac{(-8x^3y^2)^2 \cdot 2x^2y^6}{32x^8y^3}$

Numerator: $64x^6y^4 \cdot 2x^2y^6 = 128x^8y^{10}$

Denominator: $32x^8y^3$

→ $4y^7$ → matches our #4!

Brown: $\frac{3x^3y^4 \cdot 6xy^{-5}}{(3xy)^2}$

Numerator: $18x^4y^{-1}$

Denominator: $9x^2y^2$

→ $2x^{2}y^{-3}$ → matches our #6!

Perfect! All matched.

---

Final Matching:

Column 1 → Column 2 (by color):

1. $2x^5y \cdot 3x^2y = 6x^7y^2$ → Light Blue

2. $(4x^4y)^2 \cdot 2x^3y^4 = 32x^{11}y^6$ → Purple

3. $\frac{36x^9y^4}{4x^7y^3} = 9x^2y$ → Orange

4. $\frac{(2xy^5)^3}{2x^3y^8} = 4y^7$ → Pink

5. $(-5x^6y^2)^2 - 12x^{12}y^4 = 13x^{12}y^4$ → Light Green

6. $\frac{6x^{10}y^4}{3x^8y^7} = 2x^2y^{-3}$ → Brown

7. $6x^{-1}y^5 \cdot 4x^{-4}y^{-2} = 24x^{-5}y^3$ → Dark Blue

8. $(3x^{-6}y^2)^3 \cdot 2x^{10}y^{-7} = 54x^{-8}y^{-1}$ → Yellow

9. $\frac{(-2xy)^2 \cdot 10x^3y^{11}}{8x^{10}y^4} = 5x^{-5}y^9$ → Red

10. $\frac{8x^3 \cdot 12xy^7}{3x^2y^4} - 15x^2y^3 = 17x^2y^3$ → Dark Green

---

Final Answer:

1. Light Blue
2. Purple
3. Orange
4. Pink
5. Light Green
6. Brown
7. Dark Blue
8. Yellow
9. Red
10. Dark Green