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Please encourage the son to work hard learn the questions and be punctual.
Special Notes:
A. I've used the caret symbol (^) to represent exponents (e.g., x^2 means x squared).
B. I've used the asterisk (*) to represent multiplication (e.g., 2x 2 means 2 multiplied by 2).
C. I've used the division symbol (/) for division (e.g., x / z^2 means x divided by z squared).
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SOLVED BOOKLET QUESTIONS
Exercise 6.1:
Q. solve
i) Find the value of x² + (1/x²) when x + (1/x) = 7
Solution:
We can use the formula:
(x + 1/x)² = x² + 2 + 1/x²
Now x + 1/x = 7
Taking square on both sides
7² = x² + 2 + 1/x²
49 = x² + 2 + 1/x²
Simplifying
x² + (1/x²) = 47
ii) Find the value of x² + (1/x²) when x - (1/x) = 3
Solution:
We can use the formula:
(x - 1/x)² = x² - 2 + 1/x²
Now x - 1/x = 3
Taking square on both sides
3² = x² - 2 + 1/x²
9 = x² - 2 + 1/x²
Therefore, x² + (1/x²) = 11
iii) Find the value of x⁴ + (1/x⁴) when x - (1/x) = 1
Solution:
We can use the formula:
(x - 1/x)² = x² - 2 + 1/x²
Now x - 1/x = 1
Taking square on both sides we get
1² = x² - 2 + 1/x²
1 = x² - 2 + 1/x²
Therefore, x² + (1/x²) = 3
Again Taking square on both sides we get
X4 + (1/x4)+ 2 = 32
X4 + (1/x4= 9 - 2
X4 + (1/x4= 7
Exercise 6.2:
Factorizations for the given expressions:
1. 3x - 9y
2. xy + xz
3. 6ab - 14ac
4. 3m²np - 6m²n
5. 30x³ - 45xy
6. 17x²y² - 51
7. 4x³ + 3x² + 2x
8. 2p² - 4p³ + 8p
9. x³y - x²y + xy²
10. 7x⁴ - 14x²y + 21xy³
11. x²y² - xy² - xy² + xyz
12. 4x³y² - 8xy + 4xy³
13. xy⁴ - 3xy³ - 6xy²
14. x²y²z + x²yz² + xy²z²
15. 77x²y - 33xy² - 55x²y²
16. 5x⁵ + 10x⁴ + 15x³
Solutions:
Factorizations for the given expressions:
1. 3x - 9y
= 3(x - 3y)
2. xy + xz
= x(y + z)
3. 6ab - 14ac
= 2a(3b - 7c)
4. 3m²np - 6m²n
= 3m²n(p - 2)
5. 30x³ - 45xy
= 15x(2x² - 3y
6. 17x²y² - 51
= 17(x²y² - 3)
7. 4x³ + 3x² + 2x
= x(4x² + 3x + 2)
8. 2p² - 4p³ + 8p
= 2p(1 - 2p + 4)
9. x³y - x²y + xy²
= xy(x² - x + y)
10. 7x⁴ - 14x²y + 21xy³
= 7x(x³ - 2xy + 3y³)
11. x²y² - xy² - xy² + xyz
= xy(xy - y - y + z)
12. 4x³y² - 8xy + 4xy³
= 4xy(x²y - 2 + y²)
13. xy⁴ - 3xy³ - 6xy²
= xy²(y² - 3y - 6)
14. x²y²z + x²yz² + xy²z²
= xyz(xy + xz + yz)
15. 77x²y - 33xy² - 55x²y²
= 11xy(7x - 3y - 5xy)
16. 5x⁵ + 10x⁴ + 15x³
= 5x³(x² + 2x + 3)
Exercise 6.3:
Solutions:
Factorizations for the given expressions:
1. ax - by + bx - ay
Factoring by grouping:
(ax + bx) + (-by - ay)
x(a + b) - y(b + a)
x(a + b) - y(a + b)
(a + b)(x - y)
2. 2ab - 6bc - a + 3c
Factoring by grouping:
(2ab - 6bc) + (-a + 3c)
2b(a - 3c) - 1(a - 3c)
(a - 3c)(2b - 1)
3. x² + 2x - 3x - 6
Factoring by grouping:
(x² + 2x) + (-3x - 6)
x(x + 2) - 3(x + 2)
(x + 2)(x - 3)
4. x² + 5x - 2x - 10
Factoring by grouping:
(x² + 5x) + (-2x - 10)
x(x + 5) - 2(x + 5)
(x + 5)(x - 2)
5. x² - 7x + 2x - 14
Factoring by grouping:
(x² - 7x) + (2x - 14)
x(x - 7) + 2(x - 7)
(x - 7)(x + 2)
6. x² + 3x - 4x - 12
Factoring by grouping:
(x² + 3x) + (-4x - 12)
x(x + 3) - 4(x + 3)
(x + 3)(x - 4)
7. y² - 9y + 3y - 27
Factoring by grouping:
(y² - 9y) + (3y - 27)
y(y - 9) + 3(y - 9)
(y - 9)(y + 3)
8. x² - 7x - 5x + 35
Factoring by grouping:
(x² - 7x) + (-5x + 35)
x(x - 7) - 5(x - 7)
(x - 7)(x - 5)
8. x² - 7x - 5x + 35
Factoring by grouping:
(x² - 7x) + (-5x + 35)
x(x - 7) - 5(x - 7)
(x - 7)(x - 5)
9. x² - 13x - 2x + 26
Factoring by grouping:
(x² - 13x) + (-2x + 26)
x(x - 13) - 2(x - 13)
(x - 13)(x - 2)
10. x² - 8x - 4x + 32
Factoring by grouping:
(x² - 8x) + (-4x + 32)
x(x - 8) - 4(x - 8)
(x - 8)(x - 4)
11. a²(pq - rs) + b²(pq - rs)
Factoring out the
common binomial factor:
(pq - rs)(a² + b²)
12. y(y - a) - b(y - a)
Factoring out the
common binomial factor:
(y - a)(y - b)
13. a(x - y) - b(x - y)
Factoring out the
common binomial factor:
(x - y)(a - b)
14. b(x + y) + cd(x + y)
Factoring out the
common binomial factor:
(x + y)(b + cd)
Exercise 6.4:
Factorization:
1. x² + 14x + 49
It is a perfect square, factorizing it
into two factors
= (x + 7)(x + 7)
2. 9a² + 12ab + 4b²
It is a perfect square, factorizing it
into two factors
= (3a + 2b)(3a + 2b)
3. 16 + 24a + 9a²
It is a perfect square, factorizing it
into two factors
= (4 + 3a)(4 + 3a)
4. 25x² + 80xy + 64y²
It is a perfect square, factorizing it
into two factors
= (5x + 8y)(5x + 8y)
5.7a⁴ + 84a² + 252
It is a perfect square, factorizing it
into two factors
= 7(a² + 6)(a² + 6)
6. 4a² + 120a + 900
It is a perfect square, factorizing it
into two factors
= 4(a + 15)(a + 15)
7. x² - 34x + 289
It is a perfect square, factorizing it
into two factors
8. 49x² - 84x + 36
It is a perfect square, factorizing it
into two factors
9. x² - 18xy + 81y²
It is a perfect square, factorizing it
into two factors
= (x - 17)(x - 17)
= (7x - 6)(7x - 6)
= (x - 9y)(x - 9y)
10. a⁴ - 26a² + 169
It is a perfect square, factorizing it
into two factors
= (a² - 13)(a² - 13)
11. 2a² - 64a + 512
It is a perfect square, factorizing it
into two factors
= 2(a - 16)(a - 16)
12. 1 - 6a²b²c + 9a⁴b⁴c²
It is a perfect square, factorizing it
into two factors
= (1 - 3a²b²c)(1 - 3a²b²c)
13. 4x⁴ + 20x³yz + 25x²y²z²
It is a perfect square, factorizing it
into two factors
= x²(2x + 5yz)(2x + 5yz)
14. (9/16)x² + xy + (4/9)y²
It is a perfect square, factorizing it
into two factors
= ((3/4)x + (2/3)y)((3/4)x + (2/3)y)
15. (49/64)x² - 2xy + (64/49)y²
It is a perfect square, factorizing it
into two factors
= ((7/8)x - (8/7)y)((7/8)x - (8/7)y)
16. (a²/b²)x² - (2ac/bd)xy + (c²/d²)y² 18. 16x⁶ - 16x⁵ + 4x⁴
It is a perfect square, factorizing it
= 4x⁴(4x² - x + 1)
into two factors
20. a⁴b⁴x² - 2a²b²c²d²xy + c⁴d⁴y²
It is a perfect square, factorizing it
into two factors
= (a²b²x - c²d²y)(a²b²x - c²d²y)
= ((a/b)x - (c/d)y)((a/b)x - (c/d)y)
Exercise 6.5:
Factorize the following expressions:
Formula is used here (a2 – b2 ) = (a +b )(a – b )
1. 9 - x² = (3 + x)(3 - x)
4. x²y² - 64a²b² =
(xy + 8ab)(xy - 8ab)
7. 7xy² - 343x
= 7x(y² - 49)
= 7x(y + 7)(y - 7)
2. -6 + 6y²
= 6(y² - 1)
= 6(y + 1)(y - 1)
5. 16a² - 400b²
= 4(4a² - 100b²)
= 4(2a + 10b)(2a - 10b)
8. 5x³ - 45x
= 5x(x² - 9)
= 5x(x + 3)(x - 3)
3. 16x²y² - 25a²b²
= (4xy + 5ab)(4xy - 5ab)
6. a²b³ - 64a²b
= a²b(b² - 64)
= a²b(b + 8)(b - 8)
9. 11(a + b)² - 99c²
= 11[(a + b)² - 9c²]
= 11[(a + b + 3c)(a + b - 3c)]
10. 75 - 3(a - b)²
= 3[25 - (a - b)²]
= 3[(5 + a - b)(5 - a + b)]
Exercise 6.6:
Factorize the expressions
1. a² + 2ab + b² - c²
(a² + 2ab + b²) – (c²)
This is a difference of squares and perfect squares. We can factor it as follows:
(a + b)² - c² = (a + b + c)(a + b - c)
2. a² + 6ab + 9b² - 16c²
This is also a difference of squares and a perfect square. We can factor it as follows:
(a + 3b)² - 16c² = (a + 3b + 4c)(a + 3b - 4c)
3. 4a² + 4ab + b² - 9c²
This is a difference of squares and a perfect square. We can factor it as follows:
(2a + b)² - 9c² = (2a + b + 3c)(2a + b - 3c)
4. x² - 4xy + 4y² - 9c²
This is a difference of squares and a perfect square. We can factor it as follows:
(x - 2y)² - 9c² = (x - 2y + 3c)(x - 2y - 3c)
5. 9a² + 12ab + 4b² - 16c²
This is a difference of squares and a perfect square. We can factor it as follows:
(3a + 2b)² - 16c² = (3a + 2b + 4c)(3a + 2b - 4c)
Exercise 6.7:
1. Find the cube of the following:
(i) x + 4
Solution:
(x + 4)³ = x³ + 3x²(4) + 3x(4²) + 4³
= x³ + 12x² + 48x + 64
(v) 2a + b
Solution:
(2a + b)³ = (2a)³ + 3(2a)²(b) +
3(2a)(b²) + b³
= 8a³ + 12a²b + 6ab² + b³
(ix) 3x + 3y
Solution:
(3x + 3y)³ = (3x)³ + 3(3x)²(3y) +
3(3x)(3y)² + (3y)³
= 27x³ + 81x²y + 81xy² + 27y³
(ii) 2m + 1
Solution:
(2m + 1)³ = (2m)³ + 3(2m)²(1) +
3(2m)(1²) + 1³
= 8m³ + 12m² + 6m + 1
(vi) 3x + 10
Solution:
(3x + 10)³ = (3x)³ + 3(3x)²(10) +
3(3x)(10²) + 10³
= 27x³ + 270x² + 900x + 1000
(x) 7 + 2b
Solution:
(7 + 2b)³ = 7³ + 3(7)²(2b) + 3(7)(2b)²
+ (2b)³
= 343 + 294b + 84b² + 8b³
(iii) a - 2b
Solution:
(a - 2b)³ = a³ - 3a²(2b) + 3a(2b)² (2b)³
= a³ - 6a²b + 12ab² - 8b³
(vii) 2m + 3n
Solution:
(2m + 3n)³ = (2m)³ + 3(2m)²(3n) +
3(2m)(3n)² + (3n)³
= 8m³ + 36m²n + 54mn² + 27n³
(xi) 4x - 2v
Solution:
(4x - 2v)³ = (4x)³ - 3(4x)²(2v) +
3(4x)(2v)² - (2v)³
= 64x³ - 96x²v + 48xv² - 8v³
(iv) 5x - 1
Solution:
(5x - 1)³ = (5x)³ - 3(5x)²(1) + 3(5x)(1²)
- 1³
= 125x³ - 75x² + 15x - 1
(viii) 4 - 3a
Solution:
(4 - 3a)³ = 4³ - 3(4)²(3a) + 3(4)(3a)² (3a)³
= 64 - 144a + 108a² - 27a³
(xii) 5m - 4n
Solution:
(5m - 4n)³ = (5m)³ - 3(5m)²(4n) +
3(5m)(4n)² - (4n)³
= 125m³ - 300m²n + 240mn² - 64n³
We will use here the Formula
(a + b)³ = a³ + b³ + 3ab(a + b)
Q2. If x + 1/x = 8, then find the value of x³ + 1/x³.
Q5. If x - 1/x = 2, then find the value of x³ - 1/x³.
So, (x + 1/x)³ = x³ + 1/x³ + 3(x)(1/x)(x + 1/x)
We know that x + 1/x = 8, so substituting that in, we
get:
8³ = x³ + 1/x³ + 3(8)
512 = x³ + 1/x³ + 24
Therefore, x³ + 1/x³ = 512 - 24 = 488
(x + 1/x)³ = x³ + 1/x³ + 3(x)(1/x)(x + 1/x)
Using the same i Formula and approach as in question
3, we get:
2³ = x³ - 1/x³ - 3(2)
8 = x³ - 1/x³ - 6
Therefore, x³ - 1/x³ = 8 + 6 = 14
Q3. If x - 1/x = 3, then find the value of x³ - 1/x³.
We can use the same Formula (a - b)³ = a³ - b³ - 3ab(a b) to solve this.
So, (x - 1/x)³ = x³ - 1/x³ - 3(x)(1/x)(x - 1/x)
We know that x - 1/x = 3, so substituting that in, we
get:
3³ = x³ - 1/x³ - 3(3)
27 = x³ - 1/x³ - 9
Therefore, x³ - 1/x³ = 27 + 9 = 36
Q4. If x + 1/x = 7, then find the value of x³ + 1/x³.
(x + 1/x)³ = x³ + 1/x³ + 3(x)(1/x)(x + 1/x)
Using the same Formula and approach as in question 2,
we get:
7³ = x³ + 1/x³ + 3(7)
343 = x³ + 1/x³ + 21
Therefore, x³ + 1/x³ = 343 - 21 = 322
Q6. Find the cube of the following by using the formula:
(a + b)³ = a³ + 3(a²) (b) + 3(a) (b²) + b³,
(i) Cube of 13:
(13)³ = (10 + 3)³
Using the formula (a + b)³ = a³ + 3(a²) (b) + 3(a) (b²) + b³, we get:
(13)³ = 10³ + 3(10²) (3) + 3(10) (3²) + 3³
(13)³ = 1000 + 900 + 270 + 27
(13)³ = 2197
(ii) Cube of 103:
(13)³ = (100 + 3)³
Using the formula (a + b)³ = a³ + 3a²b + 3ab² + b³, we get:
(103)³ = 100³ + 3(100²)(3) + 3(100)(3²) + 3³
(103)³ = 1000000 + 90000 + 2700 + 27
(103)³ = 1092727
(iii) Cube of 0.99:
(0.99)³ = (1 - 0.01)³
Using the formula (a - b)³ = a³ - 3a²b + 3ab² - b³, we get:
0.99³ = 1³ - 3(1²)(0.01)+ 3(10.0) (1²) - (0.01)³
0.99³ = 1 - 0.03 + 0.0003 - 0.000001
0.99³ = 0.969799
Exercise 6.8
1. Write equations for the following statements:
i. The difference between father's age and daughter's age is 26 years.
Let x represent the father's age and y represent the daughter's age. The equation would be:
x - y = 26
ii. The price of 6 biscuits is equal to the price of one chocolate.
Let x represent the price of a biscuit and y represent the price of a chocolate. The equation would be:
6x = y
iii. If a number is added to three times of another number, the sum is 25.
Let x and y represent the two numbers. The equation would be:
x + 3y = 25
iv. The division of the sum of two numbers by their difference is equal to 1 (2nd number is less than 1st).
Let x and y represent the two numbers with x > y. The equation would be:
(x + y) / (x - y) = 1
v. Twice of any age increased by 7 years becomes y years.
Let A represent the age. The equation would be:
2A + 7 = y
2. Find two solutions for the equation 2x + y = 3
To find solutions, we can choose any value for x or y and solve for the other variable.
Solution 1: Let x = 0
2(0) + y = 3
y=3
Solution: (0, 3)
Solution 2: Let y = 1
2x + 1 = 3
2x = 2
x=1
Solution: (1, 1)
3. Find three solutions for the equations x + y = 2
Solution 1: Let x = 0
0+y=2
y=2
Solution: (0, 2)
Solution 2: Let y = 0
x+0=2
x=2
Solution: (2, 0)
Solution 3: Let x = 1
1+y=2
y=1
Solution: (1, 1)
4. Find four solutions for the equations y = 2x
Solution 1: Let x = 0
y = 2(0)
y=0
Solution: (0, 0)
Solution 2: Let x = 1
y = 2(1)
y=2
Solution: (1, 2)
Solution 3: Let x = 2
y = 2(2)
y=4
Solution: (2, 4)
Solution 4: Let x = -1
y = 2(-1)
y = -2
Solution: (-1, -2)
5. Is (1, 2) a solution set of x + y = 3 and 2x + 7y = 16?
To check, substitute x = 1 and y = 2 into both equations.
Equation 1:
1+2=3
Yes, it holds true.
Equation 2:
2(1) + 7(2) = 16
2 + 14 = 16
Yes, it holds true.
Therefore, (1, 2) is a solution set for both equations.
6. Which one of (3, 1) and (0, 3) is a solution of 2x + 5y = 15 and x = y - 2?
Solution 1: (3, 1)
2(3) + 5(1) = 15
6 + 5 = 15
Yes, it holds true.
3=1-2
No, it doesn't hold true.
Solution 2: (0, 3)
2(0) + 5(3) = 15
0 + 15 = 15
Yes, it holds true.
0=3-2
Yes, it holds true.
Therefore, (0, 3) is a solution for both equations
Exercise 6.9
Q1. Find the solution set by using the method of equating the coefficients Method:
Solution Method:
a. Multiply one or both equations by a suitable number so that the coefficients of one variable
become equal in magnitude but opposite in sign.
b. Add the two equations to eliminate that variable.
c. Solve the resulting equation for the remaining variable.
d. Substitute the value found in step 3 into either of the original equations to find the value of the eliminated variable.
Solutions:
(i)
x+y=2
2x + 5y = -1
Multiply the first equation by 2:
2x + 2y = 4
2x + 5y = -1
Subtract the equations 2 from equations 1, we get :
-3y = 5
y = -5/3
Substitute y = -5/3 into the first equation:
x - 5/3 = 2
x = 11/3
Solution set: {(11/3, -5/3)}
(ii)
x-y=0
x - 2y = 4
Subtract the equations 2 from equations 1, we get
y = -4
Substitute y = -4 into the first equation:
x+4=0
x = -4
Solution set: {(-4, -4)}
(iii)
2x + 3y = 3
x + 5y = 5
Multiply the second equation by 2:
2x + 3y = 3
2x + 10y = 10
Subtract the equations 2 from equations 1, we get
-7y = -7
y=1
Substitute y = 1 into the second equation:
x+5=5
x=0
Solution set: {(0, 1)}
(iv)
x - 4y = 4
4x - y = 16
Multiply the first equation by 4:
4x - 16y = 16
4x - y = 16
Subtract the equations 2 from equations 1, we get
-15y = 0
y=0
Substitute y = 0 into the first equation:
x-0=4
x=4
Solution set: {(4, 0)}
(v)
3x - 4y = 7
2x - 3y = 6
Multiply the first equation by 2, and the second equation by 3:
6x - 8y = 14
6x - 9y = 18
Subtract the equations 2 from equations 1, we get:
y = -4
Substitute y = -4 into the first equation:
3x + 16 = 7
3x = -9
x = -3
Solution set: {(-3, -4)}
(vi)
5x + y = 27
3x + 5y = 0
Multiply the first equation by 5:
25x + 5y = 135
3x + 5y = 0
Subtract the equations 2 from equations 1, we get:
22x = 135
x = 135/22
Substitute x = 135/22 into the first equation:
675/22 + y = 27
y = -33/2
Solution set: {(135/22, -33/2)}
Q2. Find the solution set by using the method of elimination by substitution:
Method:
a. Solve one of the equations for one of the variables.
b. Substitute the expression found in step 1 into the other equation.
c. Solve the resulting equation for the remaining variable.
D.Substitute the value found in step 3 into either of the original equations to find the value of the eliminated variable.
Solutions:
(i)
2x + 2y = 5
x - 2y = 3
Solve the second equation for x:
x = 2y + 3
Substitute x = 2y + 3 into the first equation:
2(2y + 3) + 2y = 5
4y + 6 + 2y = 5
6y = -1
y = -1/6
Substitute y = -1/6 into x = 2y + 3:
x = 2(-1/6) + 3
x = 8/3
Solution set: {(8/3, -1/6)}
(ii)
5x + 2y = 15
-2x + y = 4
Solve the second equation for y:
y = 2x + 4
Substitute y = 2x + 4 into the first equation:
5x + 2(2x + 4) = 15
5x + 4x + 8 = 15
9x = 7
x = 7/9
Substitute x = 7/9 into y = 2x + 4:
y = 2(7/9) + 4
y = 46/9
Solution set: {(7/9, 46/9)}
(iii)
6x + y = 2
x - 4y = 15
Solve the second equation for x:
x = 4y + 15
Substitute x = 4y + 15 into the first equation:
6(4y + 15) + y = 2
24y + 90 + y = 2
25y = -88
y = -88/25
Substitute y = -88/25 into x = 4y + 15:
x = 4(-88/25) + 15
x = 27/25
Solution set: {(27/25, -88/25)}
(iv)
2x + 7y = 10
3x + y = 3
Solve the second equation for y:
y = 3 - 3x
Substitute y = 3 - 3x into the first equation:
2x + 7(3 - 3x) = 10
2x + 21 - 21x = 10
-19x = -11
x = 11/19
Substitute x = 11/19 into y = 3 - 3x:
y = 3 - 3(11/19)
y = 36/19
Solution set: {(11/19, 36/19)}
(v)
2x - 4y = -10
y - 5x = -5
Solve the second equation for y:
y = 5x - 5
Substitute y = 5x - 5 into the first equation:
2x - 4(5x - 5) = -10
2x - 20x + 20 = -10
-18x = -30
x = 5/3
Substitute x = 5/3 into y = 5x - 5:
y = 5(5/3) - 5
y = 10/3
Solution set: {(5/3, 10/3)}
(vi)
x + 8y = 15
3x - y = 0
Solve the second equation for y:
y = 3x
Substitute y = 3x into the first equation:
x + 8(3x) = 15
x + 24x = 15
25x = 15
x = 3/5
Substitute x = 3/5 into y = 3x:
y = 3(3/5)
y = 9/5
Solution set: {(3/5, 9/5)}
Q3. Cross-Multiplication
Let the two equations be:
a₁x + b₁y + c₁ = 0 ---- (i)
a₂x + b₂y + c₂ = 0 --- (ii)
Multiplying (i) by b₂ and (ii) by b₁, we have:
a₁b₂x + b₁b₂y + b₂c₁ = 0---- (iii)
a₂b₁x + b₁b₂y + b₁c₂ = 0 ---- (iv)
Subtracting (iii) from (iv):
a₂b₁x + b₁b₂y + b₁c₂ = 0 ---- (iv)
±a₁b₂x ± b₁b₂y ± b₂c₁ = 0 ---- (iii) , we get
a₂b₁x - a₁b₂x + b₁c₂ - b₂c₁ = 0
x(a₂b₁ - a₁b₂) = b₂c₁ - b₁c₂
x = (b₂c₁ - b₁c₂) / (a₁b₂ - a₂b₁)
Now, multiplying (i) by a₂ and (ii) by a₁, we have:
a₁a₂x + a₂b₁y + a₂c₁ = 0 ----- (v)
a₁a₂x + a₁b₂y + a₁c₂ = 0 ------ (vi)
Subtracting (v) from (vi):
a₁a₂x + a₁b₂y + a₁c₂ = 0 ------ (vi)
±a₁a₂x ± a₂b₁y ± a₂c₁ = 0 ----- (v)
a₁b₂y - a₂b₁y + a₁c₂ - a₂c₁ = 0
y(a₁b₂ - a₂b₁) = a₁c₂ - a₂c₁
y = (a₁c₂ - a₂c₁) / (a₁b₂ - a₂b₁)
x / (b₁c₂ - b₂c₁) = y / (a₁c₂ - a₂c₁) = 1 / (a₁b₂ - a₂b₁)
SEE THE DIAGRAM FROM THE BOOK
________________________________________________________________________________________________
Example: Find the solution set with the method of cross multiplication.
Solution:.
2x + y - 5 = 0
3x - 4y - 2 = 0
Rewrite the given equations to have zero on the right-hand side:
2x + y - 5 = 0
3x - 4y - 2 = 0
Write down the solution using the cross-multiplication method:
=
=
Simplify the expressions:
=
Solve for x and y:
x = (-22) / (-11) = 2
y = (-11) / (-11) = 1
Therefore, the solution is x = 2 and y = 1.
Exercise 6.10:
Q1. Ahmad added 5 in the twice of a number. Then he subtracted half of the number from the result. Finally, he got the
answer 8. Find the number.
Let's denote the unknown number as "x".
Twice of the number: 2x
Adding 5:
2x + 5
Subtracting half of the number:
2x + 5 - (x/2)
Result equal to 8:
2x + 5 - (x/2) = 8
Now, solve the equation:
2x + 5 - (x/2) = 8
Multiplying both sides by 2 to eliminate the fraction:
4x + 10 - x = 16
3x + 10 = 16
3x = 6
x=2
Therefore, the number is 2.
Q2. If we add 3 in the half of a number, we get the same result as we subtract 1 from the quarter of the number. Find
the number.
let's denote the unknown number as "x".
Half of the number= x/2
Adding 3
x/2 + 3
Quarter of the number= x/4
Subtracting 1:
x/4 - 1
According to statement of question
x/2 + 3 = x/4 - 1
x/2 + 3 = x/4 - 1
Multiplying both sides by 4 to eliminate the fractions:
2x + 12 = x - 4
Subtracting x and 12 from both sides:
x = -16
Therefore, the number is -16.
Q 3: The sum of two numbers is 5 and their difference is 1. Find the numbers.
Let's denote the two numbers as "a" and "b".
We can set up two equations based on the given information:
a + b = 5 (sum)
a - b = 1 (difference)
Adding the two equations:
(a + b) + (a - b) = 5 + 1
a+b+a–b
=6
2a = 6
Dividing both sides by 2:
a=3
Substituting a = 3 into the first equation:
3+b=5
3+b–3 =5-3
b=2
Therefore, the two numbers are 3 and 2.
Q4. The difference of two numbers is 4. The sum of twice one number and 3 times the other number is 43. Find the
numbers.
Let's denote the two numbers as x and y. We can set up the following system of equations:
x - y = 4 ------ (1)
2x + 3y = 43 ------ (2)
We can solve this system of equations using the elimination method.
Multiply the first equation by 2:
2x - 2y = 8
Subtract the first equation from the second equation:
(2x + 3y) - (2x - 2y) = 43 - 8
5y = 35
y = 35 / 5
y=7
Substitute y = 7 into the first equation to find x:
x-7=4
x=4+7
x = 11
Therefore, the two numbers are 11 and 7.
Q5: Adnan is 7 years older than Adeel.Find their ages when 1/4 of Adnan's age is equal to 1/5 of Adeel's age.
Solution:
Let Adnan's age be =X
Adeel's age be= Y
Adnan is 7 years older than Adeel, which gives us the equation x = y + 7
¼ of Adnan's age is equal to 1/2
¼x=½Y
Multiply both sides of the second equation by 4 to eliminate the fractions, resulting in
x = 2y
2y = y + 7
Y=7
Now
X = y + 7 , so
X = 7+7 = 14
So, Adnan is 14 years old and Adeel is 7 years old
Q11. The numerator and denominator of a fraction are increased by 1, the fraction becomes ⅔. If the
numerator and denominator of the same fraction are decreased by 2, it becomes ⅓. Find the fraction.
Solution:
Let the original fraction be x/y.
According to the condition 1
Increasing numerator and denominator by 1:
(x+1)/(y+1) = 2/3
Cross-multiplying:
3(x+1) = 2(y+1)
3x + 3 = 2y + 2
3x – 2y = 2 – 3
3x – 2y = -1 ------------------ (1)
According to the condition 2:
Decreasing numerator and denominator by 2:
(x-2)/(y-2) = 1/3
Cross-multiplying: 3(x-2) = y-2
3x - 6 = y – 2
3x – y = -2 + 6
3x – y = 4 -------------------- (2)
Subtracting equation 2 from equation 1:
(3x – 2y) – (3x – y) = -1 – (4)
3X – 2Y – 3X + Y = -5
-y = -5
Y=5
Substituting y = 5 into equation 1:
3x – 2(5) = -1
3x = - 1 +10
3X = 9
X =3
Therefore, the original fraction is X/Y = 3/5.
Exercise 6.10:
Q1. Solve the equations to eliminate "x" using the substitution method.
i)
ax - b = 0 ---------------- (1)
cx - d = 0 --------------- (2)
From the first equation, we can express x in terms of b and a:
x = b/a
Substituting this value of x into the second equation:
c(b/a) - d = 0
Simplifying:
cb - ad = 0
ii)
2x + 3y = 5 ------------ (1)
x - y = 2 --------------- (2)
From the second equation, we can express x in terms of y:
x=y+2
Substituting this value of x into the first equation:
2(y + 2) + 3y = 5
Simplifying:
5y + 4 = 5
Solving for y:
y = 1/5
Substituting y back into the expression for x:
x = 1/5 + 2 = 11/5
iii)
x + a = b ----------------- (1)
x2 + a2 = b2 -------------- (2)
From the first equation, we can express x in terms of a and b:
x=b-a
Substituting this value of x into the second equation:
(b - a) 2 + a2 = b2
Expanding and simplifying:
b2 - 2ab + a2 + a2 = b2
2a2 - 2ab = 0
2a(a - b) = 0
This equation holds true when either a = 0 or a = b.
iv)
a - b = 2x ----------------------- (1)
a2 + b2 = 3x2 -------------------- (2)
From the first equation, we can express x in terms of a and b:
x = (a - b)/2
Substituting this value of x into the second equation:
a2 + b2 = 3((a - b)/2) 2
Expanding and simplifying:
4a2 + 4b2 = 3(a2 - 2ab + b2)
Simplifying further:
a2 + 2ab + b2 = 0
Factoring:
(a + b) 2 = 0
This equation holds true when a = -b.
v)
x - m = l ------------------------ (1)
(l - m)x + a = 0 -------------- (2)
From the first equation, we can express x in terms of m and l:
x=l+m
Substituting this value of x into the second equation:
(l - m)(l + m) + a = 0
Expanding and simplifying:
l2 - m2 + a = 0
R. Exercise 6:
Q2. Simultaneous Linear Equations
i. What are simultaneous linear equations?
Simultaneous linear equations are a set of two or more linear equations with the same set of variables. The
solution to a simultaneous linear equation is the set of values for the variables that satisfy all the equations in the
system.
ii. Write any three methods for solving simultaneous linear equations.
Three common methods for solving simultaneous linear equations are:



Elimination Method: In this method, we eliminate one variable by adding or subtracting the equations.
Substitution Method: In this method, we solve for one variable in terms of the other and substitute it into
the other equation.
Graphical Method: In this method, we graph the equations and find the point of intersection, which
represents the solution.
iii. How many equations are required for elimination of one variable?
Two equations are required for the elimination of one variable.
Q3. Find the value of x⁴ + 1/x⁴ when x + 1/x = 7.
We can use the formula
(a + b)² = a² + 2ab + b² to solve
Let x + 1/x = a. Then, a² = 7² = 49.
Expanding (a² - 2)²:
(a² - 2)² = a⁴ - 4a² + 4
Substituting a² = 49:
(49 - 2)² = a⁴ - 4(49) + 4
47² = a⁴ - 192 + 4
2209 = a⁴ - 188
a⁴ = 2397
Therefore, x⁴ + 1/x⁴ = 2397.
Q4. Factorize the following:
i. 3xy + 6x²y² + 9xz
Factor out the common factor of 3x:
3x(y + 2xy² + 3z)
ii. y⁴ - 12y² + 36
This is a perfect square:
(y² - 6)²
iii. x⁸ - y⁸
This is a difference of squares:
(x⁴ - y⁴)(x⁴ + y⁴)
(x² - y²)(x² + y²)(x⁴ + y⁴)
Q5. Find the cube of the following:
ii. 2x - 3y
Using the formula (a - b)³ = a³ - 3a²b + 3ab² - b³, we get:
(2x - 3y)³ = 8x³ - 36x²y + 54xy² - 27y³
iii. 7a - b
(7a - b)³ = 343a³ - 147a²b + 21ab² - b³
Q7. Eliminate "x" by substitution method from the following equations.
i). ax - b = 0------------ (1)
cx² + mx = 0--------- (2)
From the first equation, x = b/a. Substitute this into the second equation:
c(b/a)² + m(b/a) = 0
(cb² + mab)/a² = 0
cb² + mab = 0
ii.
lx - n = 0 ------------------------ (1)
sx² + tx + u = 0 ---------------- (2)
From the first equation, x = n/l.
Substitute this into the second equation:
s(n/l)² + t(n/l) + u = 0
(sn² + tnl + ul²)/l² = 0
sn² + tnl + ul² = 0
Answers
1.
A
2.
D
3.
C
4.
B
5.
B
6.
B
7.
A
8.
C
PART GEOMETRY
UNIT - 7
Polygons
A polygon is a closed plane figure with three or more straight sides.
Polygons are named according to the number of sides.
The most common polygons are:
1. Triangle (3 sides)
2. Quadrilateral (4 sides)
3. Pentagon (5 sides)
4. Hexagon (6 sides)
5. Heptagon (7 sides)
Parallelograms
A parallelogram is a special type of quadrilateral whose pairs of opposite sides are parallel. For example,
quadrilateral ABCD is a parallelogram because AB || DC and AD || BC.
A parallelogram has the following properties:
1. Both pairs of opposite sides are parallel.
2. Both pairs of opposite sides are congruent.
3. Both pairs of opposite angles are congruent.
4. Consecutive angles are supplementary.
Definitions and measurements for a regular pentagon, hexagon, and octagon,
Regular Pentagon:
Regular pentagon
a. A five-sided polygon with all sides and angles equal.
b. The sum of the interior angles is 540 degrees.
c. Each interior angle measures
= 108 degrees.
Regular Hexagon:
a. A six-sided polygon with all sides and angles equal.
b. The sum of the interior angles is 720 degrees.
c. Each interior angle measures
= 120 degrees.
Regular Octagon:
a. An eight-sided polygon with all sides and angles equal.
b. The sum of the interior angles is 1080 degrees.
c. Each interior angle measures
=135 degrees.
Example 4: Given that QRST is a parallelogram, find the value of x in the diagram below:
Solution:
Since opposite sides of a parallelogram are
congruent,
we have m∠(x+15) = 127° (Opposite angles in a
parallelogram).
Therefore, m∠x = 127° - 15° = 112°.
Example 5:
Given that DEFG is a parallelogram, determine the values of x and y.
Solution:
From the figure, we get m∠G = 70° + 45° = 115°.
Since m∠G + m∠D = 180° (Opposite angles in a parallelogram),
we have x + y = 180° - 115° = 65°.
Therefore, x = 65° - y.
m∠F = m∠D
m∠(7x - 5) = 65°
7x - 5 = 65°
7x = 70°
x = 10
So, we have x = 10 and y = 13.
Example 6:
Given that ABCD is a parallelogram, find the value of x.
Solution:
We know that in a parallelogram, the diagonals
bisect each other.
Thus, we get:
mDE = mBE
4x² + 5 cm = 41 cm
4x² - 41 - 5 = 0
4x² = 36
x² = 9
x=3
Examples page # 120
Example 2:
Determine the values of angles A, B, C, and D in the figure to the right where the lines p and q are parallel to each
other.
Solution:
Since ∠B is the alternate interior angle to the given angle of 75°, so m∠B
= 75°.
∠C and the given angle of 75° are corresponding angles, so m∠C = 75°.
∠A and ∠B are angles on the same side of the transversal, so:
m∠A + m∠B = 180°
m∠A + 75° = 180°
m∠A = 180° - 75° = 105°
Similarly, ∠D is an adjacent supplementary angle to the given angle of
75°, so:
m∠D + 75° = 180°
m∠D = 180° - 75° = 105°
Thus, m∠A = 105°, m∠B = 75°, m∠C = 75°, and m∠D = 105°.
Example 3:
Find the value of x, y, and z where lines a and b are parallel and lines c
and d are parallel to each other.
Solution:
Since a || b, 2x = 42° (alternate interior angles).
Therefore, m∠x = 21°.
Again, since c || d, m∠y = 42° (corresponding angles).
m∠y + m∠z = 180° (interior angles).
42° + m∠z = 180°
m∠z = 180° - 42° = 138°.
Exercise 9.2
1. The length of the sides of a triangle are 60m, 153m and 111m. Find the area of the triangle.
2. Find the area of triangles, when lengths of the sides are given below:
i. 13cm, 14cm, 15cm
ii. 5cm, 12cm, 13cm
iii. 103cm, 115cm, 13cm
3. Find the missing elements as required in each of the following with the help of Hero's formula:
i. a = 5m, b = 7m, s = 9m, c = ?, ΔABC = ?
ii. a = 10m, b = 8m, s = 12m, c = ?, ΔABC = ?
iii. a = 3m, s = 9.5m, c = 9m, b = ?
iv. a = 3.5m, b = 2.5m, c = 4.5m, ΔABC = ?
4. Find the area of the quadrilateral region ABCD. All measurements are in cm.
i. a = 19, b = 12, c = 15, d = 20, e = 23
ii. a = 12, b = 14, c = 17, d = 19, e = 21
iii. a = 2, b = 2.5, c = 3, d = 1.5, e = 3.5
iv. a = 1.7, b = 1, c = 1.3, d = 1.8, e = 2.1
5. The given figure, ABCD is of a rectangle of sides 8cm and 12cm. E and F are the mid-points of the sides BC and
AD respectively. By using Pythagoras Theorem and Hero's Formula, find:
a. The areas of the triangles ABE and FDC.
b. The area of the parallelogram AECF.
Solutions
Q1.
We need to find the area of a triangle with sides 60m, 153m, and 111m.
We can use Hero's formula to find the area.
Hero's formula states that the area of a triangle with sides a, b, and c is given by:
√
Where s = (a+b+c)/2
s = (60+153+111)/2 = 162
Now, let's substitute the values into the formula:
√
= 3240 m²
Therefore, the area of the triangle is 3240 square meters.
Q2.
We will use the same approach as in question 1 to find the areas of the triangles.
i. For the triangle with sides 13cm, 14cm, and 15cm:
s = (13+14+15)/2 = 21
√
Area = 84 cm²
ii. For the triangle with sides 5cm, 12cm, and 13cm:
s = (5+12+13)/2 = 15
Area = √
Area = 30 cm²
iii. For the triangle with sides 103cm, 115cm, and 13cm:
s = (103+115+13)/2 = 115.5
√
Area = 780 cm²
Q3.
i. We are given a = 5m, b = 7m, and s = 9m. We need to find c and ΔABC.
We can use the formula s = (a+b+c)/2 to find c:
9 = (5+7+c)/2
c = 11m
Now, we can use Hero's formula to find ΔABC:
Area = √
ΔABC = 6√3 m²
ii. We are given a = 10m, b = 8m, and s = 12m. We need to find c and ΔABC.
Using s = (a+b+c)/2, we get:
12 = (10+8+c)/2
c = 14m
ΔABC = √
ΔABC = 48 m²
iii. We are given a = 3m, s = 9.5m, and c = 9m. We need to find b.
Using s = (a+b+c)/2, we get:
9.5 = (3+b+9)/2
b = 8m
ΔABC = √
= 15 m2
iv. We are given a = 3.5m, b = 2.5m, and c = 4.5m. We need to find ΔABC.
s = (3.5+2.5+4.5)/2 = 5.25
ΔABC = √
ΔABC = 3.9375 m²
Q4.
We need to find the area of the quadrilateral ABCD.
We can divide the quadrilateral into two triangles: ABD and BCD.
We can use Hero's formula to find the area of each triangle, and then add the areas together to get the area of
the quadrilateral.
i. For triangle ABD:
s = (19+12+20)/2 = 25.5
√
Area(ABD) = 114 cm²
For triangle BCD:
s = (20+15+23)/2 = 29
√
Area (BCD) = 156 cm²
Area (ABCD) = Area (ABD) + Area (BCD) = 114 + 156 = 270 cm²
ii. Similarly, for the other quadrilaterals, we can find the areas of the triangles and add them up.
Q5.
The given figure, ABCD is of a rectangle of sides 8cm and 12cm. E and
F are the mid-points of the sides BC and AD respectively. By using
Pythagoras Theorem and Hero's Formula, find:
a. The areas of the triangles ABE and FDC.
b. The area of the parallelogram AECF.
Solution
We need to find the areas of triangles ABE and FDC, and the area of parallelogram AECF.
a. First, let's find the lengths of AE and BF:
AE = CF = 12/2 = 6 cm
Now, we can use Pythagoras theorem to find the length of BE and FD:
BE = FD = √
= 10 cm
For triangle ABE:
s = (6+8+10)/2 = 12
Area (ABE) = √
= 24 cm2
Area (ABE) = Area (FDC) = 24 cm2
(ii) Area of AECF = Area of ABCD - 2x (Area of ABE)
= 12x8 – 2x 24
= 96 – 48
= 48 cm2