5 Mark Question

Question 15 Marks

Multiply $−\frac{3}{2}\text{x}^2\text{y}^3 \ \text{by} \ (2\text{x} − \text{y})$ and verify the answer for x = 1 and y = 2.

Answer

To find the product, we will use distributive law as follows:$−\frac{3}{2}\text{x}^2\text{y}^3 \ \times \ (2\text{x} − \text{y})$
$=\big(−\frac{3}{2}\text{x}^2\text{y}^3×2\text{x}\big)−\big(−\frac{3}{2}\text{x}^2\text{y}^3×\text{y}\big)$
$=\big(−3\text{x}^{2+1}\text{y}^3\big)−\big(−\frac{3}{2}\text{x}^2\text{y}^{3+1}\big)$
$=−3\text{x}^3\text{y}^3+\frac{3}{2}\text{x}^2\text{y}^4$
Substituting x = 1 and y = 2 in the result, we get:$=−3\text{x}^3\text{y}^3+\frac{3}{2}\text{x}^2\text{y}^4$
$=−3(1)^3(2)^3+\frac{3}{2}(1)^2(2)^4$
$=−3×1×8+\frac{3}{2}×1×16$
$=−24+24$
$=0$
Thus, the product is $−3\text{x}^3\text{y}^3+\frac{3}{2}\text{x}^2\text{y}^4$ and its value for x = 1 and y = 2 is 0.

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Question 25 Marks

Express product as a monomials and verify the result for $x=1, y=2$:
$\left(-x y^3\right) \times\left(y x^3\right) \times(x y)$

Answer

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m \times a^n$$=a^{m+n}$
We have:
$\left(-x y^3\right) \times\left(y x^3\right) \times(x y)$
$=(-1) \times\left(x \times x^3 \times x\right) \times\left(y^3 \times y \times y\right)$
$=(-1) \times\left(x^{1+3+1}\right) \times\left(y^{3+1+1}\right)$
$=-x^5 y^5$
To verify the result, we substitute $x=1$ and $y=2$ in LHS; we get:
$\text { LHS }=\left(-x y^3\right) \times\left(y x^3\right) \times(x y)$
$=\left[(-1) \times 1 \times 2^3\right] \times\left(2 \times 1^{3)} \times(1 \times 2)\right.$
$=[(-1) \times 1 \times 8] \times(2 \times 1) \times 2$
$=(-8) \times 2 \times 2$
$=-32$
Substituting $x=1$ and $y=2$ in RHS, we get:
$\text { RHS }=-x^5 y^5$
$=(-1)(1)^5(2)^5$
$=(-1) \times 1 \times 32$
$=-32$
Because LHS is equal to RHS, the result is correct.
Thus, the answer is $=-x^5 y^5$

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Question 35 Marks

Subtract 3x - 4y - 7z from the sum of x - 3y + 2z and -4x + 9y - 11z.

Answer

Let first add the expressions x - 3y + 2z and -4x + 9y - 11z. We get:

(x - 3y + 2z) + (-4x +9y - 11z)
= x - 3y + 2z - 4x + 9y - 11z
= x - 4x - 3y + 9y + 2z - 11z (Collecting like terms)
= -3x + 6y - 9z (Combining like terms)
Now, subtracting the expression 3x -4y - 7z from the above sum, we get:

(-3x + 6y - 9z) - (3x - 4y - 7z)
= -3x + 6y - 9z - 3x + 4y + 7z
= -3x - 3x + 6y + 4y - 9z + 7z (Collecting like terms)
= -6x + 10y - 2z (Combining like terms)

Thus, the answer is -6x + 10y - 2z.

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Question 45 Marks

Find the following product and verify the result for x = -1, y = -2:
$\big(\frac{1}{3}\text{x}−\frac{\text{y}^2}{5}\big)\big(\frac{1}{3}\text{x}+\frac{\text{y}^2}{5}\big)$

Answer

To multiply, we will use distributive law as follows:$\big(\frac{1}{3}\text{x}−\frac{\text{y}^2}{5}\big)\big(\frac{1}{3}\text{x}+\frac{\text{y}^2}{5}\big)$
$=\Big[\frac{1}{3}\text{x}\big(\frac{1}{3}\text{x}+\frac{\text{y}^2}{5}\big)\Big]−\Big[\frac{\text{y}^2}{5}\big(\frac{1}{3}\text{x}+\frac{\text{y}^2}{5}\big)\Big]$
$=\Big[\frac{1}{9}\text{x}^2+\frac{\text{xy}^2}{15}\Big]−\Big[\frac{\text{xy}^2}{15}+\frac{\text{y}^4}{25}\Big]$
$=\frac{1}{9}\text{x}^2+\frac{\text{xy}^2}{15}−\frac{\text{xy}^2}{15}−\frac{\text{y}^4}{25}$
$=\frac{1}{9}\text{x}^2−\frac{\text{y}^4}{25}$
$\therefore\big(\frac{1}{3}\text{x}−\frac{\text{y}^2}{5}\big)\big(\frac{1}{3}\text{x}+\frac{\text{y}^2}{5}\big)=\frac{1}{9}\text{x}^2−\frac{\text{y}^4}{25}$
Now, we will put x = -1 and y = -2 on both the sides to verify the result.$\text{LHS}=\big(\frac{1}{3}\text{x}−\frac{\text{y}^2}{5}\big)\big(\frac{1}{3}\text{x}+\frac{\text{y}^2}{5}\big)$
$=\Big[\frac{1}{3}(−1)−\frac{(−2)^2}{5}\Big]\Big[\frac{1}{3}(−1)+\frac{(−2)^2}{5}\Big]$
$=\big(−\frac{1}{3}−\frac{4}{5}\big)\big(−\frac{1}{3}+\frac{4}{5}\big)$
$=\big(−\frac{17}{15}\big)\big(\frac{7}{15}\big)$
$=\frac{-119}{225}$
$\text{RHS}=\frac{1}{9}\text{x}^2−\frac{\text{y}^4}{25}$
$=\frac{1}{9}(−1)^2−\frac{(−2)^4}{25}$
$=\frac{1}{9}×1−\frac{16}{25}$
$=-\frac{119}{225}$
Because LHS is equal to RHS, the result is verified.Thus, the answer is $\frac{1}{9}\text{x}^2−\frac{\text{y}^4}{25}.$

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Question 55 Marks

Find product:
$\big(\frac{1}{8}\text{x}^2\text{y}^4\big)×\big(\frac{1}{4}\text{x}^4\text{y}^2\big)×(\text{xy})×5$

Answer

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a ^{ m } \times a ^{ n }= a ^{ m + n }$.
We have:
$\big(\frac{1}{8}\text{x}^2\text{y}^4\big)×\big(\frac{1}{4}\text{x}^4\text{y}^2\big)×(\text{xy})×5$
$=\big(\frac{1}{8}×\frac{1}{4}×5\big)×\big(\text{x}^2×\text{x}^4×\text{x}\big)×\big(\text{y}^4×\text{y}^2×\text{y}\big)$
$=\big(\frac{1}{8}×\frac{1}{4}×5\big)×\big(\text{x}^{2+4+1}\big)×\big(\text{y}^{4+2+1}\big)$
$=\frac{5}{32}\text{x}^7\text{y}^7$
To verify the result, we substitute x = 1 and y = 2 in LHS; we get:
$\text{LHS}=\big(\frac{1}{8}\text{x}^2\text{y}^4\big)×\big(\frac{1}{4}\text{x}^4\text{y}^2\big)×(\text{xy})×5$
$=\big(\frac{1}{8}×(1)^{2}×(2)^{4}\big)×\big(\frac{1}{4}×(1)^{4}×(2)^{2}\big)×(1×2)×5$
$=\big(\frac{1}{8}×(1)^{2}×(2)^{4}\big)×\big(\frac{1}{4}×1×4\big)\times(1\times2)\times5$
$=2×1×2×5$
$=20$
Substituting x = 1 and y = 2 in RHS, we get:
$\text{RHS}=\frac{5}{32}\text{x}^7\text{y}^7$
$=\frac{5}{32}(1)^7(2)^7$
$=\frac{5}{32}\times1\times128$
$=20$
Because LHS is equal to RHS, the result is correct.
Thus, the answer is $=\frac{5}{32}\text{x}^7\text{y}^7.$

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Question 65 Marks

Find the following product and verify the result for $x=-1, y=-2$ :
$\left(x^2 y-1\right)\left(3-2 x^2 y\right)$

Answer

To multiply, we will use distributive law as follows:
$\left(x^2 y-1\right)\left(3-2 x^2 y\right)$
$=x^2 y\left(3-2 x^2 y\right)-1 x\left(3-2 x^2 y\right)$
$=3 x^2 y-2 x^4 y^2-3+2 x^2 y$
$=5 x^2 y-2 x^4 y^2-3$
$\therefore\left(x^2 y-1\right)\left(3-2 x^2 y\right)=5 x^2 y-2 x^4 y^2-3$
Now, we put $x=-1$ and $y=-2$ on both sides to verify the result.
$\text { LHS }=\left(x^2 y-1\right)\left(3-2 x^2 y\right)$
$=\left[(-1)^2(-2)-1\right]\left[3-2(-1)^2(-2)\right]$
$=[1 \times(-2)-1][3-2 \times 1 \times(-2)]$
$=(-2-1)(3+4)$
$=-3 \times 7$
$\text { RHS }=5 x^2 y-2 x^4 y^2-3$
$=5(-1)^2(-2)-2(-1)^4(-2)^2-3$
$=[5 \times 1 \times(-2)]-[2 \times 1 \times 4]-3$
$=-10-8-3$
$=-21$
Because LHS is equal to RHS, the result is verified.
Thus, the answer is $5 x^2 y-2 x^4 y^2-3$

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Question 75 Marks

Simplify:
$\left(x^3-2 x^2+3 x-4\right)(x-1)-(2 x-3)\left(x^2-x+1\right)$

Answer

To simplify, we will proceed as follows: $\left(x^3-2 x^2+3 x-4\right)(x-1)-(2 x-3)\left(x^2-x+1\right)$
$=\left[\left(x^3-2 x^2+3 x-4\right)(x-1)\right]-[(2 x-3)(x 2-x+1)]$
$=\left[x\left(x^3-2 x^2+3 x-4\right)-1\left(x^3-2 x^2+3 x-4\right)\right]-\left[2 x\left(x^2-x+1\right)-3\left(x^2-x+1\right)\right] \text { (Distributive law) }$
$=x^4-2 x^3+3 x^2-4 x-x 3+2 x^2-3 x+4-\left[2 x^3-2 x^2+2 x-3 x^2+3 x-3\right]$
$=x^4-2 x^3+3 x^2-4 x-x 3+2 x^2-3 x+4-2 x^3+2 x^2-2 x+3 x^2-3 x+3$
$=x 4-2 x^3-2 x^3-x^3+3 x^2+2 x^2+2 x^2+3 x^2-4 x-3 x-2 x-3 x+4+3 \text { (Rearranging) }$
$=x^4-5 x^3+10 x^2-12 x+7 \text { (Combining like terms) }$
Thus, the answer is $x^4-5 x^3+10 x^2-12 x+7$.

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Question 85 Marks

Write down the product of $-8 x^2 y^6$ and $-20 x y$. Verify the product for $x=2.5, y=1$.

Answer

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a ^{ m } \times$
$a^n=a^{m+n} .$
We have:
$\left(-8 x^2 y^6\right) \times(-20 x y)$
$=[(-8) \times(-20)] \times\left(x^2 \times x\right) \times\left(y^6 \times y\right)$
$=[(-8) \times(-20)] \times\left(x^{2+1}\right) \times\left(y^{6+1}\right)$
$=-160 x^3 y^7$
$\therefore\left(-8 x^2 y^6\right) \times(-20 x y)=-160 x^3 y^7$
Substituting $x=2.5$ and $y=1$ in LHS, we get:
$\left(-8 x^2 y^6\right) \times(-20 x y)$
$=\left[-8(2.5)^2(1)^6\right] \times[-20(2.5)(1)]$
$=[-8(6.25)(1)] \times[-20(2.5)(1)]$
$=(-50) \times(-50)$
$=2500$
Substituting $x=2.5$ and $y=1$ in RHS, we get:
$\text { RHS }=-160 x^3 y^7$
$=-160(2.5)^3(1)^7$
$=-160(15.625) \times 1$
$=-2500$
Because LHS is equal to RHS, the result is correct.
Thus, the answer is $-160 x^3 y^7$.

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Question 95 Marks

If $\text{x}−\frac{1}{\text{x}}=3,$ find the values of $\text{x}^2+\frac{1}{\text{x}^2} $and $\text{x}^4+\frac{1}{\text{x}^4}.$

Answer

Let us consider the following equation:$\text{x}−\frac{1}{\text{x}}=3,$
Squaring both sides, we get:
$\big(\text{x}−\frac{1}{\text{x}}\big)^2=(3)^2=9$
$\big(\text{x}−\frac{1}{\text{x}}\big)^2=9$
$⇒\text{x}^2−2×\text{x}×\frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2=9$
$⇒\text{x}^2−2+\frac{1}{\text{x}^2}=9$
$⇒\text{x}^2+\frac{1}{\text{x}^2}=11$
Squaring both sides again, we get:
$\big(\text{x}^2+\frac{1}{\text{x}^2}\big)^2=(11)^2=121$
$\big(\text{x}^2+\frac{1}{\text{x}^2}\big)^2=121$
$⇒\big(\text{x}^2)^2+2\big(\text{x}^2\big)\big(\frac{1}{\text{x}^2}\big)+\big(\frac{1}{\text{x}^2}\big)^2=121$
$⇒\text{x}^4+2+\frac{1}{\text{x}^4}=121$
$⇒\text{x}^4+\frac{1}{\text{x}^4}=119$

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Question 105 Marks

If $2 x+3 y=14$ and $2 x-3 y=2$, find the value of $x y$.
[Hint: Use $(2 x+3 y)^2-(2 x-3 y)^2=24 x y$ ]

Answer

We will use the identity $(a+b)(a-b)=a^2-b^2 a+b a-b=a^2-b^2$ to obtain the value of $x y$.
Squaring $(2 x+3 y)$ and $(2 x-3 y)$ both and then subtracting them, we get:
$\left.(2 x+3 y)^2-(2 x-3 y)^2=\{(2 x+3 y)+(2 x-3 y)\}(2 x+3 y)-(2 x-3 y)\right\}=4 x \times 6 y=24 x y$
$\Rightarrow(2 x+3 y)^2-(2 x-3 y)^2=24 x y$
$\Rightarrow 24 x y=(2 x+3 y)^2-(2 x-3 y)^2$
$\Rightarrow 24 x y=(14)^2-(2)^2$
$\Rightarrow 24 x y=(14+2)(14-2)\left(\because(a+b)(a-b)=a^2-b^2\right)$
$\Rightarrow 24 x y=16 \times 12$
$\left.\Rightarrow x y=\frac{16 \times 12}{24} \text { (Dividing both sides by } 24\right)$
$\Rightarrow x y=8$

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Question 115 Marks

If $\text{x}^2+\frac{1}{\text{x}^2}=18,$ find the values of $\text{x}+\frac{1}{\text{x}}$ and $\text{x}-\frac{1}{\text{x}}.$

Answer

Let us consider the following equation:
$\text{x}+\frac{1}{\text{x}}$
Squaring both sides, we get:
$\big(\text{x}+\frac{1}{\text{x}}\big)^2=\text{x}^2+2×\text{x}×\frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2\\=\text{x}^2-2+\frac{1}{\text{x}^2} $
$\Rightarrow\big(\text{x}+\frac{1}{\text{x}}\big)^2=\text{x}^2+2+\frac{1}{\text{x}^2}$
$\Rightarrow\big(\text{x}+\frac{1}{\text{x}}\big)^2=20$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\pm\sqrt{20}$ (Taking square root of both sides)
Now, let us consider the following expression:
$\text{x}-\frac{1}{\text{x}}$
Squaring the above expression, we get:
$\big(\text{x}-\frac{1}{\text{x}}\big)^2=\text{x}^2-2×\text{x}×\frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2\\=\text{x}^2-2+\frac{1}{\text{x}^2 }$
$\Rightarrow\big(\text{x}-\frac{1}{\text{x}}\big)^2=\text{x}^2-2+\frac{1}{\text{x}^2 }$ $\big[(\text{a}−\text{b}\big)^2=\text{a}^2+\text{b}^2−2\text{ab}\big]$
$\Rightarrow\big(\text{x}-\frac{1}{\text{x}}\big)^2=16$ $\big(\because \text{x}^2+\frac{1}{\text{x}^2}=18\big)$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\pm4$ (Taking square root of both sides)

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Question 125 Marks

Find the values of the following expressions:
If $\text{x}+\frac{1}{\text{x}}=9,$ find the value of $\text{x}^4+\frac{1}{\text{x}^4}.$

Answer

Let us consider the following equation:
$\text{x}+\frac{1}{\text{x}}=9$
Squaring both sides, we get:
$\big(\text{x}+\frac{1}{\text{x}}\big)^2=(9)^2=81$
$\Rightarrow\big(\text{x}+\frac{1}{\text{x}}\big)^2=81$
$⇒\text{x}^2+2×\text{x}×\frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2=81$
$⇒\text{x}^2+2+\frac{1}{\text{x}^2}=81$
$⇒\text{x}^2+\frac{1}{\text{x}^2}=79$ (Subtracting 2 from both sides)
Now, squaring both sides again, we get:
$\big(\text{x}^2+\frac{1}{\text{x}^2}\big)^2=(79)^2=6241$
$\Rightarrow\big(\text{x}^2+\frac{1}{\text{x}^2}\big)^2=6241$
$⇒\big(\text{x}^2\big)^2+2\big(\text{x}^2\big)\big(\frac{1}{\text{x}^2}\big)+\big(\frac{1}{\text{x}^2}\big)^2=6241$
$⇒\text{x}^4+2+\frac{1}{\text{x}^4}=6241$
$⇒\text{x}^4+\frac{1}{\text{x}^4}=6239$

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Question 135 Marks

If $\text{x}+\frac{1}{\text{x}}=12,$ find the value of $\text{x}−\frac{1}{\text{x}}.$

Answer

Let us consider the following equation:
$\text{x}+\frac{1}{\text{x}}=12$
Squaring both sides, we get:
$\big(\text{x}+\frac{1}{\text{x}}\big)^2=(12)^2=144$
$\Rightarrow\big(\text{x}+\frac{1}{\text{x}}\big)^2=144$
$⇒\text{x}^2+2×\text{x}×\frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2=144 $ $\big[ (\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}\big] $
$⇒\text{x}^2+2+\frac{1}{\text{x}^2}=144$
$⇒\text{x}^2+\frac{1}{\text{x}^2}=142$ (Subtracting 2 from both sides)
Now
$\big(\text{x}−\frac{1}{\text{x}}\big)^2=\text{x}^2−2×\text{x}×\frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2\\=\text{x}^2−2+\frac{1}{\text{x}^2 } $ $\big[ (\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}\big] $
$\big(\text{x}−\frac{1}{\text{x}}\big)^2=\text{x}^2−2+\frac{1}{\text{x}^2 } $
$\big(\text{x}−\frac{1}{\text{x}}\big)^2=142-2$ $\big(\because \text{x}^2+\frac{1}{\text{x}^2}=142\big)$
$\big(\text{x}−\frac{1}{\text{x}}\big)^2=140$
$\text{x}−\frac{1}{\text{x}}=\pm\sqrt{140}$ (Taking square root)

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Question 145 Marks

Write the following squares of binomials as trinomials:
$\big(\frac{\text{x}}{\text{y}}−\frac{\text{y}}{\text{x}})^2$

Answer

We will use the identities $(a+b)^2=a^2+2 a b+b^2$ and $(a-b)^2=a^2-2 a b+b^2$ to convert the squares of binomials as trinomials.
$\big(\frac{\text{x}}{\text{y}}−\frac{\text{y}}{\text{x}})^2$
$=\big(\frac{\text{x}}{\text{y}})^2−2\big(\frac{\text{x}}{\text{y}}\big)\big(\frac{\text{y}}{\text{x}}\big)+\big(\frac{\text{y}}{\text{x}})^2$
$=\frac{\text{x}^2}{\text{y}^2}−2+\frac{\text{y}^2}{\text{x}^2}$

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Maths 5 Mark Question

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