Algebraic Expressions

algebraic expressions class 8 selina

Table of Contents

Exercise: 11-A

Q1: Multiple Choice Type

i. The sum of \(a-b+ab,\ b-c+bc\) and \(c-a+ca\) is:

Step 1: Write the expression:
\((a – b + ab) + (b – c + bc) + (c – a + ca)\)
Step 2: Rearranging and grouping like terms:
\(a – b + ab + b – c + bc + c – a + ca\)
Step 3: Group same variables:
\((a – a) + (-b + b) + (-c + c) + ab + bc + ca\)
Step 4: Simplify:
\(0 + 0 + 0 + ab + bc + ca = ab + bc + ca\)
Answer: c. \(ab + bc + ca\)

ii. \(\left(x^3-5x^2+7\right)+\left(3x^2+5x-2\right)+\left(2x^3-x+7\right)\) is equal to:

Step 1: Group like terms:
\(x^3 + 2x^3 = 3x^3\)
\(-5x^2 + 3x^2 = -2x^2\)
\(5x – x = 4x\)
\(7 – 2 + 7 = 12\)
Step 2: Final expression:
\(3x^3 – 2x^2 + 4x + 12\)
Answer: a. \(3x^3 – 2x^2 + 4x + 12\)

iii. \(\left(x^3-5x^2+3x+2\right)-\left(6x^2-4x^3+3x+5\right)\) is equal to:

Step 1: Distribute negative sign:
= \(x^3 – 5x^2 + 3x + 2 – 6x^2 + 4x^3 – 3x – 5\)
Step 2: Group like terms:
\(x^3 + 4x^3 = 5x^3\)
\(-5x^2 – 6x^2 = -11x^2\)
\(3x – 3x = 0\)
\(2 – 5 = -3\)
Step 3: Final expression:
\(5x^3 – 11x^2 -3\)
Answer: c. \(5x^3 – 11x^2 – 3\)

iv. \(p-\left(p-q\right)-q-\left(q-p\right)\) is equal to:

Step 1: Remove brackets carefully:
= \(p – p + q – q – q + p\)
Step 2: Simplify:
\((p – p) + (q – q) + (-q) + p = 0 + 0 – q + p\)
= \(p – q\)
Answer: b. \(p – q\)

v. \(\left(ab-bc\right)-\left(ca-bc\right)+\left(ca-ab\right)\) is:

Step 1: Expand the terms:
= \(ab – bc – ca + bc + ca – ab\)
Step 2: Group like terms:
\(ab – ab = 0\), \(-bc + bc = 0\), \(-ca + ca = 0\)
Step 3: Final expression:
= 0
Answer: d. 0


Q2: Separate the constants and variables from the following:
\(-7,\ 7+x,\ 7x+yz,\ \sqrt5,\ \sqrt{xy},\ \frac{3yz}{8},\ 4.5y-3x,\ 8-5,\ 8-5x,\ 8x-5y\times p\ and\ 3y^2z\div4x\)

i. Identify constants:

Step 1: A constant is a term that has no variable — just a number.
Step 2: Go through each expression:
– \(-7\) → constant
– \(7 + x\) → contains variable \(x\)
– \(7x + yz\) → variables \(x, y, z\)
– \(\sqrt{5}\) → constant (since 5 is a number)
– \(\sqrt{xy}\) → contains variables \(x, y\)
– \(\frac{3yz}{8}\) → variables \(y, z\)
– \(4.5y – 3x\) → variables \(x, y\)
– \(8 – 5\) → constant = 3
– \(8 – 5x\) → variable \(x\)
– \(8x – 5y \times p\) → variables \(x, y, p\)
– \(3y^2z \div 4x\) → variables \(x, y, z\)
Answer:Constants: \(-7,\ \sqrt{5},\ 8 – 5 \)

ii. Identify expressions containing variables:

Step 1: List remaining expressions that include variables:
– \(7 + x\) → variable \(x\)
– \(7x + yz\) → variables \(x, y, z\)
– \(\sqrt{xy}\) → variables \(x, y\)
– \(\frac{3yz}{8}\) → variables \(y, z\)
– \(4.5y – 3x\) → variables \(x, y\)
– \(8 – 5x\) → variable \(x\)
– \(8x – 5y \times p\) → variables \(x, y, p\)
– \(3y^2z \div 4x\) → variables \(x, y, z\)
Answer:Expressions with variables: \(7 + x,\ 7x + yz,\ \sqrt{xy},\ \frac{3yz}{8},\ 4.5y – 3x,\ 8 – 5x,\ 8x – 5y \times p,\ 3y^2z \div 4x\)


Q3: Write the number of terms in each of the following polynomials:

i. \(5x^2 + 3 \times a x\)

Step 1: Simplify the expression:
\(5x^2 + 3ax\)
Step 2: Count individual terms:
→ \(5x^2\) is one term
→ \(3ax\) is another term
Answer: 2 terms

ii. \(x \div 4 – 7\)

Step 1: Simplify the expression:
\(\frac{x}{4} – 7\)
Step 2: Count the terms:
→ \(\frac{x}{4}\) is one term
→ \(-7\) is another term
Answer: 2 terms

iii. \(ax – by + y \times z\)

Step 1: Simplify the expression:
\(ax – by + yz\)
Step 2: Count the terms:
→ \(ax\) is one term
→ \(-by\) is another term
→ \(yz\) is the third term
Answer: 3 terms

iv. \(23 + a \times b \div 2\)

Step 1: Simplify the expression:
\(23 + \frac{ab}{2}\)
Step 2: Count the terms:
→ \(23\) is one term
→ \(\frac{ab}{2}\) is another term
Answer: 2 terms


Q4: Separate monomials, binomials, trinomials and multinomial from the following algebraic expressions:
\(8-3x,\ xy^2,\ 3y^2-5y+8,\ 9x-3x^2+15x^3-7,\ 3x\times5y,\ 3x\div5y,\ 2y\div7+3x-7\ and\ 4-ax^2+bx+y\)

Monomials (1 term):
xy²
3x × 5y
(3x ÷ 5y) = 3x/5y
Binomials (2 terms):
8 – 3x
Trinomials (3 terms):
3y² – 5y + 8
(2y ÷ 7 + 3x – 7)
Polynomials (more than 3 terms):
8 – 3x
3y² – 5y + 8
(2y ÷ 7 + 3x – 7)
9x – 3x² + 15x³ – 7
4 – ax² + bx + y
Answer:
Monomials: xy², 3x × 5y, 3x ÷ 5y
Binomials: 8 – 3x
Trinomials: 3y² – 5y + 8, 2y ÷ 7 + 3x – 7
Multinomials: 8 – 3x, 3y² – 5y + 8, 2y ÷ 7 + 3x – 7, 9x – 3x² + 15x³ – 7, 4 – ax² + bx + y



Q5: Write the degree of each polynomial given below:

i. \(xy + 7z\)

Explanation: Degree of term \(xy = 1+1 = 2\), term \(7z = 1\).
Answer: 2

ii. \(x^2 – 6x^3 + 8\)

Explanation: Highest power of x is in \(x^3\).
Answer: 3

iii. \(y – 6y^2 + 5y^8\)

Explanation: Highest power of y is in \(y^8\).
Answer: 8

iv. \(xyz – 3\)

Explanation: Degree of \(xyz = 1+1+1 = 3\).
Answer: 3

v. \(xy + yz^2 – zx^3\)

Explanation: Degrees: \(xy = 2\), \(yz^2 = 1+2=3\), \(zx^3 = 1+3 = 4\)
Answer: 4

vi. \(x^5y^7 – 8x^3y^8 + 10x^4y^4z^4\)

Explanation: Degrees:
\(x^5y^7 = 12\),
\(x^3y^8 = 11\),
\(x^4y^4z^4 = 12\)
Maximum degree = 12
Answer: 12


Q6: Write the coefficient of:

i. ab in 7abx

Answer: 7x

ii. 7a in 7abx

Answer: bx

iii. \(5x^2\) in \(5x^2 – 5x\)

Answer: 1

iv. 8 in \(a^2 – 8ax + a\)

Answer: -ax

v. \(4xy\) in \(x^2 – 4xy + y^2\)

Answer: -1


Q7: Evaluate:

i. \(-7x^2+18x^2+3x^2-5x^2\)

Step 1: Combine like terms of \(x^2\):
\(-7x^2 + 18x^2 + 3x^2 – 5x^2 = (18 – 7 + 3 – 5)x^2\)
Step 2: Simplify the coefficients:
\((18 – 7 + 3 – 5) = 9\)
Answer: \(9x^2\)

ii. \(b^2y – 9b^2y + 2b^2y – 5b^2y\)

Step 1: Combine like terms of \(b^2y\):
\((1 – 9 + 2 – 5)b^2y\)
Step 2: Simplify the coefficients:
\((1 – 9 + 2 – 5) = -11\)
Answer: \(-11b^2y\)

iii. \(abx – 15abx – 10abx + 32abx\)

Step 1: Combine like terms of \(abx\):
\((1 – 15 – 10 + 32)abx\)
Step 2: Simplify the coefficients:
\((1 – 15 – 10 + 32) = 8\)
Answer: \(8abx\)

iv. \(7x – 9y + 3 – 3x – 5y + 8\)

Step 1: Group like terms:
\((7x – 3x) + (-9y – 5y) + (3 + 8)\)
Step 2: Simplify each group:
\(4x – 14y + 11\)
Answer: \(4x – 14y + 11\)

v. \(3x^2 + 5xy – 4y^2 + x^2 – 8xy – 5y^2\)

Step 1: Group like terms:
\((3x^2 + x^2) + (5xy – 8xy) + (-4y^2 – 5y^2)\)
Step 2: Simplify each group:
\(4x^2 – 3xy – 9y^2\)
Answer: \(4x^2 – 3xy – 9y^2\)


Q8: Add:

i. \(5a + 3b,\ a – 2b,\ 3a + 5b\)

Step 1: Write all expressions:
\(5a + 3b + a – 2b + 3a + 5b\)
Step 2: Group like terms:
\((5a + a + 3a) + (3b – 2b + 5b)\)
Step 3: Simplify:
\(9a + 6b\)
Answer: \(9a + 6b\)

ii. \(8x – 3y + 7z,\ -4x + 5y – 4z,\ -x – y – 2z\)

Step 1: Combine all expressions:
\(8x – 3y + 7z – 4x + 5y – 4z – x – y – 2z\)
Step 2: Group like terms:
\((8x – 4x – x) + (-3y + 5y – y) + (7z – 4z – 2z)\)
Step 3: Simplify each group:
\(3x + 1y + 1z = 3x + y + z\)
Answer: \(3x + y + z\)

iii. \(3b – 7c + 10,\ 5c – 2b – 15,\ 15 + 12c + b\)

Step 1: Add all expressions:
\(3b – 7c + 10 + 5c – 2b – 15 + 15 + 12c + b\)
Step 2: Group like terms:
\((3b – 2b + b) + (-7c + 5c + 12c) + (10 – 15 + 15)\)
Step 3: Simplify:
\(2b + 10c + 10\)
Answer: \(2b + 10c + 10\)

iv. \(a – 3b + 3,\ 2a + 5 – 3c,\ 6c – 15 + 6b\)

Step 1: Add all expressions:
\(a – 3b + 3 + 2a + 5 – 3c + 6c – 15 + 6b\)
Step 2: Group like terms:
\((a + 2a) + (-3b + 6b) + (-3c + 6c) + (3 + 5 – 15)\)
Step 3: Simplify:
\(3a + 3b + 3c – 7\)
Answer: \(3a + 3b + 3c – 7\)

v. \(13ab – 9cd – xy,\ 5xy,\ 15cd – 7ab,\ 6xy – 3cd\)

Step 1: Combine all expressions:
\(13ab – 9cd – xy + 5xy + 15cd – 7ab + 6xy – 3cd\)
Step 2: Group like terms:
\((13ab – 7ab) + (-9cd + 15cd – 3cd) + (-xy + 5xy + 6xy)\)
Step 3: Simplify:
\(6ab + 3cd + 10xy\)
Answer: \(6ab + 3cd + 10xy\)

vi. \(x^3 – x^2y + 5xy^2 + y^3,\ -x^3 – 9xy^2 + y^3,\ 3x^2y + 9xy^2\)

Step 1: Combine all terms:
\(x^3 – x^2y + 5xy^2 + y^3 – x^3 – 9xy^2 + y^3 + 3x^2y + 9xy^2\)
Step 2: Group like terms:
\((x^3 – x^3) + (-x^2y + 3x^2y) + (5xy^2 – 9xy^2 + 9xy^2) + (y^3 + y^3)\)
Step 3: Simplify:
\(0 + 2x^2y + 5xy^2 + 2y^3\)
Answer: \(2x^2y + 5xy^2 + 2y^3\)


Q9: Find the total savings of a boy who saves ₹\(\left(4x-6y\right)\), ₹\(\left(6x+2y\right)\), ₹\(\left(4y-x\right)\) and ₹\(\left(y-2x\right)\) in four consecutive weeks.

Expression: Add the four expressions:
(4x − 6y), (6x + 2y), (4y − x), and (y − 2x)
Step 1: Write all the expressions to be added:
(4x − 6y) + (6x + 2y) + (4y − x) + (y − 2x)
Step 2: Group like terms together:
= (4x + 6x − x − 2x) + (−6y + 2y + 4y + y)
Step 3: Add the coefficients of like terms:
= (4 + 6 − 1 − 2)x + (−6 + 2 + 4 + 1)y
= (7)x + (1)y
Step 4: Simplify:
= 7x + y
Answer: The total savings = ₹(7x + y)


Q10: Subtract:

i. \(4xy^2\) from \(3xy^2\)

Required Expression: \(3xy^2 – 4xy^2\)
Answer: \(-xy^2\)

ii. \(-2x^2y + 3xy^2\) from \(8x^2y\)

Required Expression: \(8x^2y – (-2x^2y + 3xy^2)\)
= \(8x^2y + 2x^2y – 3xy^2\)
= \(10x^2y – 3xy^2\)
Answer: \(10x^2y – 3xy^2\)

iii. \(3a – 5b + c + 2d\) from \(7a – 3b + c – 2d\)

Required Expression: \((7a – 3b + c – 2d) – (3a – 5b + c + 2d)\)
= \(7a – 3b + c – 2d – 3a + 5b – c – 2d\)
= \((7a – 3a) + (-3b + 5b) + (c – c) + (-2d – 2d)\)
= \(4a + 2b – 4d\)
Answer: \(4a + 2b – 4d\)

iv. \(x^3 – 4x – 1\) from \(3x^3 – x^2 + 6\)

Required Expression: \((3x^3 – x^2 + 6) – (x^3 – 4x – 1)\)
= \(3x^3 – x^2 + 6 – x^3 + 4x + 1\)
= \((3x^3 – x^3) – x^2 + 4x + (6 + 1)\)
= \(2x^3 – x^2 + 4x + 7\)
Answer: \(2x^3 – x^2 + 4x + 7\)

v. \(6a + 3\) from \(a^3 – 3a^2 + 4a + 1\)

Required Expression: \((a^3 – 3a^2 + 4a + 1) – (6a + 3)\)
= \(a^3 – 3a^2 + 4a + 1 – 6a – 3\)
= \(a^3 – 3a^2 – 2a – 2\)
Answer: \(a^3 – 3a^2 – 2a – 2\)

vi. \(cab – 4cad – cbd\) from \(3abc + 5bcd – cda\)

Required Expression: \((3abc + 5bcd – cda) – (cab – 4cad – cbd)\)
= \(3abc + 5bcd – cda – cab + 4cad + cbd\)
= \(3abc – abc + 5bcd + cbd – cda + 4cad\)
= \(2abc + 6bcd + 3cda\)
Answer: \(2abc + 6bcd – 3cda\)

vii. \(a^2 + ab + b^2\) from \(4a^2 – 3ab + 2b^2\)

Required Expression: \((4a^2 – 3ab + 2b^2) – (a^2 + ab + b^2)\)
= \(4a^2 – 3ab + 2b^2 – a^2 – ab – b^2\)
= \((4a^2 – a^2) + (-3ab – ab) + (2b^2 – b^2)\)
= \(3a^2 – 4ab + b^2\)
Answer: \(3a^2 – 4ab + b^2\)


Q11:

i. Take away \(-3x^3+4x^2-5x+6\) from \(3x^3-4x^2+5x-6\)

Step 1: Write the expression to subtract.
= \([3x^3 – 4x^2 + 5x – 6] – [-3x^3 + 4x^2 – 5x + 6]\)
Step 2: Remove brackets and change signs of second polynomial.
= \(3x^3 – 4x^2 + 5x – 6 + 3x^3 – 4x^2 + 5x – 6\)
Step 3: Combine like terms.
= \((3x^3 + 3x^3) + (-4x^2 – 4x^2) + (5x + 5x) + (-6 – 6)\)
= \(6x^3 – 8x^2 + 10x – 12\)
Answer: \(6x^3 – 8x^2 + 10x – 12\)

ii. Take \(m^2+m+4\) from \(-m^2+3m+6\) and the result from \(m^2+m+1\).

Step 1: First subtract the two polynomials.
= \([-m^2 + 3m + 6] – [m^2 + m + 4]\)
Step 2: Change signs of second polynomial.
= \(-m^2 + 3m + 6 – m^2 – m – 4\)
Step 3: Combine like terms.
= \(-2m^2 + 2m + 2\)
Step 4: Now subtract from \(m^2 + m + 1\)
= \([m^2 + m + 1] – [-2m^2 + 2m + 2]\)
Step 5: Change signs.
= \(m^2 + m + 1 + 2m^2 – 2m – 2\)
Step 6: Combine like terms.
= \((m^2 + 2m^2) + (m – 2m) + (1 – 2)\)
= \(3m^2 – m – 1\)
Answer: \(3m^2 – m – 1\)


Q12: Subtract the sum of \(5y^2+y-3\) and \(y^2-3y+7\) to obtain \(6y^2+y-2\)?

Step 1: Find the sum of two polynomials.
= \([5y^2 + y – 3] + [y^2 – 3y + 7]\)
= \(6y^2 – 2y + 4\)
Step 2: Subtract the sum from given expression:
= \([6y^2 + y – 2] – [6y^2 – 2y + 4]\)
Step 3: Change signs.
= \(6y^2 + y – 2 – 6y^2 + 2y – 4\)
Step 4: Combine like terms.
= \(0y^2 + 3y – 6\)
= \(3y – 6\)
Answer: \(3y – 6\)


Q13: What must be added to \(x^4-x^3+x^2+x+3\) to obtain \(x^4+x^2-1\)?

Step 1: Let the required polynomial be \(P(x)\)
Then, \(x^4 – x^3 + x^2 + x + 3 + P(x) = x^4 + x^2 – 1\)
Step 2: Transpose given expression.
\(P(x) = [x^4 + x^2 – 1] – [x^4 – x^3 + x^2 + x + 3]\)
Step 3: Change signs of second polynomial.
= \(x^4 + x^2 – 1 – x^4 + x^3 – x^2 – x – 3\)
Step 4: Combine like terms.
= \(0 + x^3 + 0 – x – 4\)
= \(x^3 – x – 4\)
Answer: \(x^3 – x – 4\)


Q14:

i. How much more than \(2x^2+4xy+2y^2\) is \(5x^2+10xy-y^2\)?

Step 1: Subtract the smaller from larger.
= \([5x^2 + 10xy – y^2] – [2x^2 + 4xy + 2y^2]\)
Step 2: Change signs of second polynomial.
= \(5x^2 + 10xy – y^2 – 2x^2 – 4xy – 2y^2\)
Step 3: Combine like terms.
= \(3x^2 + 6xy – 3y^2\)
Answer: \(3x^2 + 6xy – 3y^2\)

ii. How much less \(2a^2+1\) is than \(3a^2-6\)?

Step 1: Subtract the smaller from the larger.
= \([3a^2 – 6] – [2a^2 + 1]\)
Step 2: Change signs.
= \(3a^2 – 6 – 2a^2 – 1\)
Step 3: Combine like terms.
= \(a^2 – 7\)
Answer: \(a^2 – 7\)


Q15: If \(x=6a+8b+9c;y=2b-3a-6c\) and \(z=c-b+3a\); find:

i. \(x + y + z\)

Step 1: Write expressions for x, y, and z.
  \(x = 6a + 8b + 9c\)
  \(y = 2b – 3a – 6c\)
  \(z = c – b + 3a\)
Step 2: Add them: \(x + y + z\)
  = \((6a + 8b + 9c) + (2b – 3a – 6c) + (c – b + 3a)\)
Step 3: Combine like terms:
a-terms: \(6a – 3a + 3a = 6a\)
  b-terms: \(8b + 2b – b = 9b\)
  c-terms: \(9c – 6c + c = 4c\)
Answer: x + y + z = 6a + 9b + 4c

ii. \(x – y + z\)

Step 1: Use expressions for z and y:
\(x = 6a + 8b + 9c\)
  \(z = c – b + 3a\)
  \(y = 2b – 3a – 6c\)
Step 2: Substitute: \(x – y + z = (6a + 8b + 9c) – (2b – 3a – 6c) + (c – b + 3a)\)
Step 3: Remove brackets (change signs carefully):
  = \(6a + 8b + 9c – 2b + 3a + 6c + c – b + 3a\)
Step 4: Combine like terms:
  a-terms: \(6a + 3a + 3a = 12a\)
  b-terms: \(8b – 2b – b = 5b\)
  c-terms: \(9c + 6c + c = 16c\)
Answer: x – y + z = 12a – 5b + 16c

iii. \(2x – y – 3z\)

Step 1: Use expressions:
\(x = 6a + 8b + 9c\)
  \(z = c – b + 3a\)
  \(y = 2b – 3a – 6c\)
Step 2: Find \(2x – y – 3z\)
  = \(2(6a + 8b + 9c) – (2b – 3a – 6c) – 3(c – b + 3a)\)
Step 3: Expand each:
  = \(12a + 16b + 18c – 2b + 3a + 6c – 3c + 3b – 9a\)
Step 4: Combine like terms:
  a-terms: \(12a + 3a – 9a = 6a\)
  b-terms: \(16b – 2b + 3b = 17b\)
  c-terms: \(18c + 6c – 3c = 21c\)
Answer: 2x – y – 3z = 6a + 17b + 21c

iv. \(3y – 2z – 5x\)

Step 1: Use expressions for x, y, z:
  \(x = 6a + 8b + 9c\)
  \(y = 2b – 3a – 6c\)
  \(z = c – b + 3a\)
Step 2: Expand each:
  \(3y = 6b – 9a – 18c\)
  \(2z = 2c – 2b + 6a\)
  \(5x = 30a + 40b + 45c\)
Step 3: Combine: \(3y – 2z – 5x\)
  = \((6b – 9a – 18c) – (2c – 2b + 6a) – (30a + 40b + 45c)\)
Step 4: Simplify:
  = \(6b – 9a – 18c – 2c + 2b – 6a – 30a – 40b – 45c\)
Step 5: Combine like terms:
  a-terms: \(-9a – 6a – 30a = -45a\)
  b-terms: \(6b + 2b – 40b = -32b\)
  c-terms: \(-18c – 2c – 45c = -65c\)
Answer: 3y – 2z – 5x = -45a – 32b – 65c


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