Answered

[tex]\[ x y(x-y) \][/tex]

[tex]\[ -3 p q(2 p+5 q) \][/tex]

h) [tex]\( 2 x y \)[/tex]

n) [tex]\( -4 x\left(x^2-2 x+1\right) \)[/tex]

k) [tex]\( -2 b c(7 b-3 c) \)[/tex]

n) [tex]\( -5 a\left(3 a^2+3 a-5\right) \)[/tex]

l) [tex]\( -5 x y \)[/tex]

o) [tex]\( -3 x y \)[/tex]

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1. If [tex]\( x=a+5 \)[/tex] and [tex]\( y=2a \)[/tex], show that [tex]\( xy=2a^2+10a \)[/tex].

2. If [tex]\( p=5x \)[/tex] and [tex]\( q=3x-1 \)[/tex], show that [tex]\( 2pq=30x^2-10x \)[/tex].

3. If [tex]\( a=x-3, b=3x \)[/tex] and [tex]\( x=2 \)[/tex], find the value of [tex]\( 4ab \)[/tex].

4. If [tex]\( m=n=2a \)[/tex] and [tex]\( a=3 \)[/tex], find the value of [tex]\( \sqrt{mn} \)[/tex].

5. If [tex]\( p=2x, q=3x \)[/tex] and [tex]\( x=3 \)[/tex], find the value of [tex]\( \sqrt{6pq} \)[/tex].



Answer :

GinnyAnswer
Sure, let's solve each question step-by-step.

### Question 1
If [tex]\( x = a + 5 \)[/tex] and [tex]\( y = 2a \)[/tex], show that [tex]\( xy = 2a^2 + 10a \)[/tex].

1. First, substitute the expressions for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = a + 5 \quad \text{and} \quad y = 2a \][/tex]

2. Now find [tex]\( xy \)[/tex]:
[tex]\[ xy = (a + 5) \cdot 2a \][/tex]

3. Distribute [tex]\( y \)[/tex] through [tex]\( x \)[/tex]:
[tex]\[ xy = 2a \cdot (a + 5) = 2a \cdot a + 2a \cdot 5 = 2a^2 + 10a \][/tex]

Therefore, [tex]\( xy = 2a^2 + 10a \)[/tex], which is what we were asked to show.

### Question 2
If [tex]\( p = 5x \)[/tex] and [tex]\( q = 3x - 1 \)[/tex], show that [tex]\( 2pq = 30x^2 - 10x \)[/tex].

1. Substitute the given expressions:
[tex]\[ p = 5x \quad \text{and} \quad q = 3x - 1 \][/tex]

2. Now compute [tex]\( pq \)[/tex]:
[tex]\[ pq = (5x) \cdot (3x - 1) \][/tex]

3. Distribute [tex]\( p \)[/tex] through [tex]\( q \)[/tex]:
[tex]\[ pq = 5x \cdot 3x - 5x \cdot 1 = 15x^2 - 5x \][/tex]

4. Now, multiply this result by 2:
[tex]\[ 2pq = 2 \cdot (15x^2 - 5x) = 30x^2 - 10x \][/tex]

So, [tex]\( 2pq = 30x^2 - 10x \)[/tex].

### Question 3
If [tex]\( a = x - 3 \)[/tex], [tex]\( b = 3x \)[/tex], and [tex]\( x = 2 \)[/tex], find the value of [tex]\( 4ab \)[/tex].

1. Substitute [tex]\( x \)[/tex] into the expressions for [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ a = 2 - 3 = -1 \quad \text{and} \quad b = 3 \cdot 2 = 6 \][/tex]

2. Compute [tex]\( 4ab \)[/tex]:
[tex]\[ 4ab = 4 \cdot (-1) \cdot 6 = -24 \][/tex]

So, the value of [tex]\( 4ab = -24 \)[/tex].

### Question 4
If [tex]\( m = n = 2a \)[/tex] and [tex]\( a = 3 \)[/tex], find the value of [tex]\( \sqrt{mn} \)[/tex].

1. Substitute [tex]\( a \)[/tex] into the expressions for [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
[tex]\[ m = 2 \cdot 3 = 6 \quad \text{and} \quad n = 2 \cdot 3 = 6 \][/tex]

2. Compute [tex]\( mn \)[/tex]:
[tex]\[ mn = 6 \cdot 6 = 36 \][/tex]

3. Find the square root of [tex]\( mn \)[/tex]:
[tex]\[ \sqrt{mn} = \sqrt{36} = 6.0 \][/tex]

So, the value of [tex]\( \sqrt{mn} = 6.0 \)[/tex].

### Question 5
If [tex]\( p = 2x \)[/tex], [tex]\( q = 3x \)[/tex], and [tex]\( x = 3 \)[/tex], find the value of [tex]\( \sqrt{6pq} \)[/tex].

1. Substitute [tex]\( x \)[/tex] into the expressions for [tex]\( p \)[/tex] and [tex]\( q \)[/tex]:
[tex]\[ p = 2 \cdot 3 = 6 \quad \text{and} \quad q = 3 \cdot 3 = 9 \][/tex]

2. Compute [tex]\( pq \)[/tex]:
[tex]\[ pq = 6 \cdot 9 = 54 \][/tex]

3. Now multiply by 6 and then find the square root:
[tex]\[ 6pq = 6 \cdot 54 = 324 \quad \text{and} \quad \sqrt{6pq} = \sqrt{324} = 18.0 \][/tex]

So, the value of [tex]\( \sqrt{6pq} = 18.0 \)[/tex].