Question 11

We use the distributive property of multiplication. This property helps us multiply. It works for addition and subtraction.

Step 1 — Understand the property

Let us understand the distributive property. For addition, it states: $a \times (b + c) = a \times b + a \times c$. The multiplier can be on the right side. It also states: $(b + c) \times a = b \times a + c \times a$. For subtraction, it states: $a \times (b - c) = a \times b - a \times c$. The multiplier can be on the right side. It also states: $(b - c) \times a = b \times a - c \times a$. We will use these rules to fill the blanks.

Step 2 — Solve part (c)

The given expression is $3 \times (5 + 8) = 3 \times 5 \ \Box\ 3 \times \underline{\ \ \ }$. This matches $a \times (b + c) = a \times b + a \times c$. Here, $a = 3$, $b = 5$, and $c = 8$. So, the box must be a plus sign. The blank must be the number $c$. The blank is 8.

$3 \times (5 + 8) = 3 \times 5 \ \mathbf{+}\ 3 \times \mathbf{8}$

$\boxed{\text{Box: + , Blank: 8}}$

Step 3 — Solve part (d)

The given expression is $(9 + 2) \times 4 = 9 \times 4 \ \Box\ 2 \times \underline{\ \ \ }$. This matches $(b + c) \times a = b \times a + c \times a$. Here, $b = 9$, $c = 2$, and $a = 4$. So, the box must be a plus sign. The blank must be the number $a$. The blank is 4.

$(9 + 2) \times 4 = 9 \times 4 \ \mathbf{+}\ 2 \times \mathbf{4}$

$\boxed{\text{Box: + , Blank: 4}}$

Step 4 — Solve part (e)

The given expression is $3 \times (\underline{\ \ \ } + 4) = 3 \ \underline{\ \ \ } + \underline{\ \ \ }$. This matches $a \times (b + c) = a \times b + a \times c$. Here, $a = 3$ and $c = 4$. Let us choose a number for $b$. Let $b = 5$. So, the first blank on the left is 5. The first blank on the right is $b$. The first blank on the right is 5. The second blank on the right is $a \times c$. This is $3 \times 4 = 12$. The second blank on the right is 12.

$3 \times (\mathbf{5} + 4) = 3 \times \mathbf{5} + \mathbf{12}$

$\boxed{\text{Blanks: 5, 5, 12}}$

Step 5 — Solve part (f)

The given expression is $(\underline{\ \ \ } + 6) \times 4 = 13 \times 4 + \underline{\ \ \ }$. This matches $(b + c) \times a = b \times a + c \times a$. Here, $c = 6$ and $a = 4$. We see $b \times a = 13 \times 4$. Since $a = 4$, $b$ must be 13. So, the first blank on the left is 13. The blank on the right is $c \times a$. This is $6 \times 4 = 24$. The blank on the right is 24.

$(\mathbf{13} + 6) \times 4 = 13 \times 4 + \mathbf{24}$

$\boxed{\text{Blanks: 13, 24}}$

Step 6 — Solve part (g)

The given expression is $3 \times (\underline{\ \ \ } + \underline{\ \ \ }) = 3 \times 5 + 3 \times 2$. This matches $a \times (b + c) = a \times b + a \times c$. Here, $a = 3$. We see $a \times b = 3 \times 5$, so $b = 5$. We see $a \times c = 3 \times 2$, so $c = 2$. The blanks on the left are 5 and 2.

$3 \times (\mathbf{5} + \mathbf{2}) = 3 \times 5 + 3 \times 2$

$\boxed{\text{Blanks: 5, 2}}$

Step 7 — Solve part (h)

The given expression is $(\underline{\ \ \ } + \underline{\ \ \ }) \times \underline{\ \ \ } = 2 \times 4 + 3 \times 4$. This matches $(b + c) \times a = b \times a + c \times a$. We see $b \times a = 2 \times 4$, so $b = 2$ and $a = 4$. We see $c \times a = 3 \times 4$, so $c = 3$ and $a = 4$. The blanks on the left are 2, 3, and 4.

$(\mathbf{2} + \mathbf{3}) \times \mathbf{4} = 2 \times 4 + 3 \times 4$

$\boxed{\text{Blanks: 2, 3, 4}}$

Step 8 — Solve part (i)

The given expression is $5 \times (9 - 2) = 5 \times 9 - 5 \times \underline{\ \ \ }$. This matches $a \times (b - c) = a \times b - a \times c$. Here, $a = 5$, $b = 9$, and $c = 2$. The blank must be the number $c$. The blank is 2.

$5 \times (9 - 2) = 5 \times 9 - 5 \times \mathbf{2}$

$\boxed{\text{Blank: 2}}$

Step 9 — Solve part (j)

The given expression is $(5 - 2) \times 7 = 5 \times 7 - 2 \times \underline{\ \ \ }$. This matches $(b - c) \times a = b \times a - c \times a$. Here, $b = 5$, $c = 2$, and $a = 7$. The blank must be the number $a$. The blank is 7.

$(5 - 2) \times 7 = 5 \times 7 - 2 \times \mathbf{7}$

$\boxed{\text{Blank: 7}}$

Step 10 — Solve part (k)

The given expression is $5 \times (8 - 3) = 5 \times 8 \ \Box\ 5 \times \underline{\ \ \ }$. This matches $a \times (b - c) = a \times b - a \times c$. Here, $a = 5$, $b = 8$, and $c = 3$. So, the box must be a minus sign. The blank must be the number $c$. The blank is 3.

$5 \times (8 - 3) = 5 \times 8 \ \mathbf{-}\ 5 \times \mathbf{3}$

$\boxed{\text{Box: - , Blank: 3}}$

Step 11 — Solve part (l)

The given expression is $(8 - 3) \times 7 = 8 \times 7 \ \Box\ 3 \times 7$. This matches $(b - c) \times a = b \times a - c \times a$. Here, $b = 8$, $c = 3$, and $a = 7$. So, the box must be a minus sign.

$(8 - 3) \times 7 = 8 \times 7 \ \mathbf{-}\ 3 \times 7$

$\boxed{\text{Box: -}}$

Step 12 — Solve part (m)

The given expression is $5 \times (12 - \underline{\ \ \ }) = \underline{\ \ \ } \ \Box\ 5 \times \underline{\ \ \ }$. This matches $a \times (b - c) = a \times b - a \times c$. Here, $a = 5$ and $b = 12$. Let us choose a number for $c$. Let $c = 3$. So, the first blank on the left is 3. The first blank on the right is $a \times b$. This is $5 \times 12 = 60$. The first blank on the right is 60. The box must be a minus sign. The box is -. The second blank on the right is $a \times c$. This is $5 \times 3 = 15$. The second blank on the right is 3. (It's $c$, not $a \times c$).

$5 \times (12 - \mathbf{3}) = \mathbf{60} \ \mathbf{-}\ 5 \times \mathbf{3}$

$\boxed{\text{Blanks: 3, 60, 3, Box: -}}$

Step 13 — Solve part (n)

The given expression is $(15 - \underline{\ \ \ }) \times 7 = \underline{\ \ \ } \ \Box\ 6 \times 7$. This matches $(b - c) \times a = b \times a - c \times a$. Here, $b = 15$ and $a = 7$. We see $c \times a = 6 \times 7$. Since $a = 7$, $c$ must be 6. So, the first blank on the left is 6. The first blank on the right is $b \times a$. This is $15 \times 7 = 105$. The first blank on the right is 105. The box must be a minus sign. The box is -.

$(15 - \mathbf{6}) \times 7 = \mathbf{105} \ \mathbf{-}\ 6 \times 7$

$\boxed{\text{Blanks: 6, 105, Box: -}}$

Step 14 — Solve part (o)

The given expression is $5 \times (\underline{\ \ \ } - \underline{\ \ \ }) = 5 \times 9 - 5 \times 4$. This matches $a \times (b - c) = a \times b - a \times c$. Here, $a = 5$. We see $a \times b = 5 \times 9$, so $b = 9$. We see $a \times c = 5 \times 4$, so $c = 4$. The blanks on the left are 9 and 4.

$5 \times (\mathbf{9} - \mathbf{4}) = 5 \times 9 - 5 \times 4$

$\boxed{\text{Blanks: 9, 4}}$

Step 15 — Solve part (p)

The given expression is $(\underline{\ \ \ } - \underline{\ \ \ }) \times \underline{\ \ \ } = 17 \times 7 - 9 \times 7$. This matches $(b - c) \times a = b \times a - c \times a$. We see $b \times a = 17 \times 7$, so $b = 17$ and $a = 7$. We see $c \times a = 9 \times 7$, so $c = 9$ and $a = 7$. The blanks on the left are 17, 9, and 7.

$(\mathbf{17} - \mathbf{9}) \times \mathbf{7} = 17 \times 7 - 9 \times 7$

$\boxed{\text{Blanks: 17, 9, 7}}$

Answer

(c) Box: +, Blank: 8 (d) Box: +, Blank: 4 (e) Blanks: 5, 5, 12 (Example choice for the first blank is 5) (f) Blanks: 13, 24 (g) Blanks: 5, 2 (h) Blanks: 2, 3, 4 (i) Blank: 2 (j) Blank: 7 (k) Box: -, Blank: 3 (l) Box: - (m) Blanks: 3, 60, 3, Box: - (Example choice for the first blank is 3) (n) Blanks: 6, 105, Box: - (o) Blanks: 9, 4 (p) Blanks: 17, 9, 7