CBSE Class 7 Mathematics Algebraic Expressions Assignment Set 04

Question 1. From the sum of \( 2q^2 + 3qr - r^2 \), \( -q^2 - qr - r^2 \) and \( qr + 2r^2 \), subtract the sum of \( 3q^2 - r^2 \) and \( -q^2 + qr + r^2 \)
Answer: First, find the sum of the first three expressions:
\( (2q^2 + 3qr - r^2) + (-q^2 - qr - r^2) + (qr + 2r^2) \)
Group like terms together:
\( = (2q^2 - q^2) + (3qr - qr + qr) + (-r^2 - r^2 + 2r^2) \)
\( = q^2 + 3qr + 0 \)
\( = q^2 + 3qr \)

Next, find the sum of the remaining two expressions:
\( (3q^2 - r^2) + (-q^2 + qr + r^2) \)
Group like terms together:
\( = (3q^2 - q^2) + qr + (-r^2 + r^2) \)
\( = 2q^2 + qr + 0 \)
\( = 2q^2 + qr \)

Now, subtract the second sum from the first sum:
\( (q^2 + 3qr) - (2q^2 + qr) \)
\( = q^2 + 3qr - 2q^2 - qr \)
\( = (q^2 - 2q^2) + (3qr - qr) \)
\( = -q^2 + 2qr \)
In simple words: Add the first three sets of terms. Then add the second two sets. Finally, take away the second answer from the first answer.
Exam Tip: When subtracting an algebraic expression, remember to change the sign of every term inside the bracket.

Question 2. Simplify combining like terms:
(i) \( 3x^2 - 5x + 7 \); \( -8x^2 + 4x - 3 \) and \( 11x^2 + 7x + 8 \)
(ii) \( 2x^2 - \frac{x}{2} + \frac{1}{3} \); \( -\frac{3x^2}{5} - \frac{1}{3}x + \frac{4}{5} \); and \( 3x^2 - 2x + 4 \)
Answer:
(i) Combine the like terms together:
\( = (3x^2 - 8x^2 + 11x^2) + (-5x + 4x + 7x) + (7 - 3 + 8) \)
\( = (3 - 8 + 11)x^2 + (-5 + 4 + 7)x + (7 - 3 + 8) \)
\( = 6x^2 + 6x + 12 \)

(ii) Group the like terms:
\( = \left(2x^2 - \frac{3}{5}x^2 + 3x^2\right) + \left(-\frac{1}{2}x - \frac{1}{3}x - 2x\right) + \left(\frac{1}{3} + \frac{4}{5} + 4\right) \)
Calculate the coefficients for each term:
For \( x^2 \) terms:
\( 2 - \frac{3}{5} + 3 = 5 - \frac{3}{5} = \frac{25 - 3}{5} = \frac{22}{5} \)
For \( x \) terms:
\( -\frac{1}{2} - \frac{1}{3} - 2 = \frac{-3 - 2 - 12}{6} = -\frac{17}{6} \)
For constant terms:
\( \frac{1}{3} + \frac{4}{5} + 4 = \frac{5 + 12 + 60}{15} = \frac{77}{15} \)
Putting them together gives:
\( = \frac{22x^2}{5} - \frac{17x}{6} + \frac{77}{15} \)
In simple words: Put matching letters and powers in their own groups. Then add or subtract the numbers in front of them.

Exam Tip: Take the lowest common multiple (LCM) of denominators carefully when combining fraction coefficients.

Question 3. Subtract \( 2x^2 + 4x - 5 \) from \( 9x^2 + 11x - 11 \)
Answer: Set up the subtraction with the expression after "from" placed first:
\( (9x^2 + 11x - 11) - (2x^2 + 4x - 5) \)
Change the sign of each term inside the second bracket:
\( = 9x^2 + 11x - 11 - 2x^2 - 4x + 5 \)
Group like terms:
\( = (9x^2 - 2x^2) + (11x - 4x) + (-11 + 5) \)
\( = 7x^2 + 7x - 6 \)
In simple words: Write the second expression first. Then switch all the plus and minus signs of the other expression and combine them.

Exam Tip: In subtraction problems, always write the expression following the word "from" at the front.

Question 4. Subtract \( \frac{2}{3}x^2y + \frac{1}{2}xy^2 + 5 \) from \( \frac{5}{6}x^2y - 3xy^2 - 11 \)
Answer: Place the term after "from" first:
\( \left(\frac{5}{6}x^2y - 3xy^2 - 11\right) - \left(\frac{2}{3}x^2y + \frac{1}{2}xy^2 + 5\right) \)
Open the brackets by reversing the signs of the second expression:
\( = \frac{5}{6}x^2y - 3xy^2 - 11 - \frac{2}{3}x^2y - \frac{1}{2}xy^2 - 5 \)
Collect the like terms:
\( = \left(\frac{5}{6} - \frac{2}{3}\right)x^2y + \left(-3 - \frac{1}{2}\right)xy^2 + (-11 - 5) \)
Simplify the coefficients:
\( \frac{5}{6} - \frac{2}{3} = \frac{5 - 4}{6} = \frac{1}{6} \)
\( -3 - \frac{1}{2} = \frac{-6 - 1}{2} = -\frac{7}{2} \)
\( -11 - 5 = -16 \)
Thus, the result is:
\( = \frac{x^2y}{6} - \frac{7xy^2}{2} - 16 \)
In simple words: Match terms with the same combination of powers, then subtract their fractional numbers step by step.

Exam Tip: Note that \( x^2y \) and \( xy^2 \) are different terms and cannot be added or subtracted together.

Question 5. Multiply
(i) \( 2x^2y^3 \) and \( 7xy^3 \)
(ii) \( \frac{2}{5}a^2b^2c \) and \( \frac{5}{2}abc \)
Answer:
(i) Multiply the coefficients and add the powers of like variables:
\( = (2 \times 7) \times (x^2 \times x) \times (y^3 \times y^3) \)
\( = 14 \times x^{2+1} \times y^{3+3} \)
\( = 14x^3y^6 \)

(ii) Multiply the numerical factors and the variable factors:
\( = \left(\frac{2}{5} \times \frac{5}{2}\right) \times (a^2 \times a) \times (b^2 \times b) \times (c \times c) \)
\( = 1 \times a^{2+1} \times b^{2+1} \times c^{1+1} \)
\( = a^3b^3c^2 \)
In simple words: Multiply the numbers normally. When multiplying the same letters, add their powers.

Exam Tip: If a variable does not show an exponent, remember that its power is 1.

Question 6. Multiply \( -4xy^2 \), \( \frac{2}{3}xz^2 \), \( \frac{4}{5}xyz \) and \( xy^2z \). Verify your result for \( x = 1 \), \( y = 2 \) and \( z = 3 \)
Answer: First, multiply the numerical coefficients and like variables together:
Product \( = \left(-4 \times \frac{2}{3} \times \frac{4}{5} \times 1\right) \times (x \cdot x \cdot x \cdot x) \times (y^2 \cdot y \cdot y^2) \times (z^2 \cdot z \cdot z) \)
\( = -\frac{32}{15} \times x^{1+1+1+1} \times y^{2+1+2} \times z^{2+1+1} \)
\( = -\frac{32}{15}x^4y^5z^4 \)

Verification for \( x = 1 \), \( y = 2 \), and \( z = 3 \):
L.H.S. \( = (-4xy^2) \times \left(\frac{2}{3}xz^2\right) \times \left(\frac{4}{5}xyz\right) \times (xy^2z) \)
Substitute the given values into each factor:
\( -4(1)(2)^2 = -4(4) = -16 \)
\( \frac{2}{3}(1)(3)^2 = \frac{2}{3}(9) = 6 \)
\( \frac{4}{5}(1)(2)(3) = \frac{24}{5} \)
\( (1)(2)^2(3) = 1 \times 4 \times 3 = 12 \)
L.H.S. \( = (-16) \times 6 \times \frac{24}{5} \times 12 \)
\( = -1152 \times \frac{24}{5} \)
\( = -\frac{27648}{5} \)

R.H.S. \( = -\frac{32}{15}x^4y^5z^4 \)
Substitute \( x = 1 \), \( y = 2 \), and \( z = 3 \):
\( = -\frac{32}{15} \times (1)^4 \times (2)^5 \times (3)^4 \)
\( = -\frac{32}{15} \times 1 \times 32 \times 81 \)
Simplify by dividing 81 and 15 by 3:
\( = -\frac{32 \times 32 \times 27}{5} \)
\( = -\frac{1024 \times 27}{5} \)
\( = -\frac{27648}{5} \)

Since L.H.S. = R.H.S., the result is verified.
In simple words: Multiply all four terms into a single answer. Then replace the letters with their numbers to make sure both sides give the same final value.

Exam Tip: Always show separate steps for L.H.S. and R.H.S. to ensure full credit on verification questions.

Question 7. Multiply \( 4xyz - 2xy \) by \( -\frac{3}{5}xy \)
Answer: Multiply each term inside the bracket by \( -\frac{3}{5}xy \):
\( = (4xyz - 2xy) \times \left(-\frac{3}{5}xy\right) \)
\( = 4xyz \times \left(-\frac{3}{5}xy\right) - 2xy \times \left(-\frac{3}{5}xy\right) \)
\( = \left(4 \times -\frac{3}{5}\right)(x \cdot x)(y \cdot y)(z) - \left(2 \times -\frac{3}{5}\right)(x \cdot x)(y \cdot y) \)
\( = -\frac{12}{5}x^2y^2z + \frac{6}{5}x^2y^2 \)
In simple words: Multiply each term inside the brackets by the single term outside, paying attention to the signs.

Exam Tip: Be careful with signs: multiplying a negative term by another negative term results in a positive term.

Question 8. Find the value of,
(i) \( 6a^3 \times -7a^2b^2 \times 10ab \) at \( a = 1, b = \frac{1}{2} \)
(ii) \( \frac{9}{2}x^2y \times \frac{1}{2}xy^2 \times \frac{3}{4}xy \) at \( x = -1, y = -1 \)
Answer:
(i) Multiply the terms together first:
\( = [6 \times (-7) \times 10] \times (a^3 \cdot a^2 \cdot a) \times (b^2 \cdot b) \)
\( = -420 a^{3+2+1} b^{2+1} \)
\( = -420 a^6 b^3 \)
Now substitute \( a = 1 \) and \( b = \frac{1}{2} \):
\( = -420 \times (1)^6 \times \left(\frac{1}{2}\right)^3 \)
\( = -420 \times 1 \times \frac{1}{8} \)
\( = -\frac{420}{8} \)
Divide numerator and denominator by 4:
\( = -\frac{105}{2} \)

(ii) Multiply the expressions first:
\( = \left(\frac{9}{2} \times \frac{1}{2} \times \frac{3}{4}\right) \times (x^2 \cdot x \cdot x) \times (y \cdot y^2 \cdot y) \)
\( = \frac{27}{16} x^{2+1+1} y^{1+2+1} \)
\( = \frac{27}{16} x^4 y^4 \)
Now substitute \( x = -1 \) and \( y = -1 \):
\( = \frac{27}{16} \times (-1)^4 \times (-1)^4 \)
Since \( (-1)^4 = 1 \):
\( = \frac{27}{16} \times 1 \times 1 \)
\( = \frac{27}{16} \)
In simple words: Simplify the multiplication into one short term before placing the numbers in for the letters.

Exam Tip: Any negative number raised to an even power becomes positive.

Question 9. Find the value of the expression \( 25x^2 + 9y^2 + 30xy \) at \( x = 8 \) and \( y = 10 \)
Answer: Rearrange the expression to identify the algebraic identity:
\( 25x^2 + 30xy + 9y^2 \)
This can be written as:
\( (5x)^2 + 2(5x)(3y) + (3y)^2 \)
Using the identity \( a^2 + 2ab + b^2 = (a + b)^2 \):
\( = (5x + 3y)^2 \)
Now substitute \( x = 8 \) and \( y = 10 \):
\( = [5(8) + 3(10)]^2 \)
\( = (40 + 30)^2 \)
\( = 70^2 \)
\( = 4900 \)
In simple words: Turn the three terms into the square of \( (5x + 3y) \). Then put in the numbers, add them, and square the sum.

Exam Tip: Factoring the expression into a perfect square first makes the arithmetic much faster and prevents errors.

Question 10. Find the value of
(i) \( 3x - 5y - x + 2y \) at \( x = 3, y = -1 \)
(ii) \( 3x^2y + 2xy - xy^2 \) at \( x = 3, y = 5 \)
Answer:
(i) Simplify by combining like terms:
\( = (3x - x) + (-5y + 2y) \)
\( = 2x - 3y \)
Now substitute \( x = 3 \) and \( y = -1 \):
\( = 2(3) - 3(-1) \)
\( = 6 + 3 \)
\( = 9 \)

(ii) Substitute \( x = 3 \) and \( y = 5 \) directly into the expression:
\( 3x^2y + 2xy - xy^2 \)
\( = 3(3)^2(5) + 2(3)(5) - (3)(5)^2 \)
\( = 3(9)(5) + 30 - 3(25) \)
\( = 135 + 30 - 75 \)
\( = 165 - 75 \)
\( = 90 \)
In simple words: Combine like terms where you can, then swap each letter with its given number and solve.

Exam Tip: Always work out powers first before multiplying the coefficients.