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 $y = 2$.
AnswerTo 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 $0$.
View full question & answer→Question 25 Marks
Express product as a monomials and verify the result for $x=1, y=2:\left(-x y^3\right) \times\left(y x^3\right) \times(x y)$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m× a^n= a^{m+n}$.
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:
$LHS$ $= (-xy^3) \times (yx^3) \times (xy) $
$= [(-1) \times 1\times 2^3] \times (2 \times 1^3)\times (1 \times 2) $
$= [(-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:
$RHS$ $= -x^5y^5$
$= (-1)(1)^5(2)^5$
$= (-1) \times 1 \times 32$
$= -32$
Because $LHS$ is equal to $RHS$, the result is correct. Thus, the answer is $ = -x^5y^5$
View full question & answer→Question 35 Marks
Subtract $3x - 4y - 7z$ from the sum of $x - 3y + 2z$ and $-4x + 9y - 11z.$
AnswerLet 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 $-6x + 10y - 2z.$
View full question & answer→Question 45 Marks
Find the following product and verify the result for $x = -1, y = -2:$
$\big(\frac{1}{3}\text{x}−\frac{\text{y}^2}{5}\big)\big(\frac{1}{3}\text{x}+\frac{\text{y}^2}{5}\big)$
AnswerTo 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 $\frac{1}{9}\text{x}^2−\frac{\text{y}^4}{25}.$
View full question & answer→Question 55 Marks
Find product: $\big(\frac{1}{8}\text{x}^2\text{y}^4\big)×\big(\frac{1}{4}\text{x}^4\text{y}^2\big)×(\text{xy})×5$
AnswerTo 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 $=\frac{5}{32}\text{x}^7\text{y}^7.$
View full question & answer→Question 65 Marks
Find the following product and verify the result for $x = -1, y = -2: \left(x^2 y-1\right)\left(3-2 x^2 y\right)$
AnswerTo 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 \times \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 $
View full question & answer→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) $
AnswerTo 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 $x^4-5 x^3+10 x^2-12 x+7$.
View full question & answer→Question 85 Marks
Write down the product of $-8x^2y^6$ and $-20xy$. Verify the product for $x = 2.5, y = 1.$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^m× a^n= a^{m+n}$.
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 $-160x^3y^7$.
View full question & answer→Question 95 Marks
If $\text{x}−\frac{1}{\text{x}}=3,$ find the values of $\text{x}^2+\frac{1}{\text{x}^2} $and $\text{x}^4+\frac{1}{\text{x}^4}.$
AnswerLet 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:
$\big(\text{x}^2+\frac{1}{\text{x}^2}\big)^2=(11)^2=121$
$\big(\text{x}^2+\frac{1}{\text{x}^2}\big)^2=121$
$⇒\big(\text{x}^2)^2+2\big(\text{x}^2\big)\big(\frac{1}{\text{x}^2}\big)+\big(\frac{1}{\text{x}^2}\big)^2=121$
$⇒\text{x}^4+2+\frac{1}{\text{x}^4}=121$
$⇒\text{x}^4+\frac{1}{\text{x}^4}=119$
View full question & answer→Question 105 Marks
If $2x + 3y = 14$ and $2x - 3y = 2,$ find the value of $xy$. $[$Hint: Use $(2x + 3y)^2- (2x - 3y)^2= 24xy]$
AnswerWe 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 $xy$.
Squaring $(2x + 3y)$ and $(2x - 3y)$ both and then subtracting them, we get:
$ (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)\}=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$
$\Rightarrow 24xy = (14 + 2)(14 - 2) (\because (a + b)(a - b) = a^2- b^2)$
$\Rightarrow 24xy = 16 \times 12$
$ \Rightarrow \text{xy}=\frac{16\times 12}{24} $ (Dividing both sides by $24$)
$\Rightarrow \text{xy}=8$
View full question & answer→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)
View full question & answer→Question 125 Marks
Find the values of the following expressions: If $\text{x}+\frac{1}{\text{x}}=9,$ find the value of $\text{x}^4+\frac{1}{\text{x}^4}.$
AnswerLet us consider the following equation: $\text{x}+\frac{1}{\text{x}}=9$ Squaring both sides,
we get: $\big(\text{x}+\frac{1}{\text{x}}\big)^2=(9)^2=81$
$\Rightarrow\big(\text{x}+\frac{1}{\text{x}}\big)^2=81$
$\Rightarrow \text{x}^2+2\times \text{x}\times \frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2=81$
$\Rightarrow \text{x}^2+2+\frac{1}{\text{x}^2}=81$
$\Rightarrow \text{x}^2+\frac{1}{\text{x}^2}=79$ (Subtracting $2$ from both sides) Now, squaring both sides again, we get: $\big(\text{x}^2+\frac{1}{\text{x}^2}\big)^2=(79)^2=6241$
$\Rightarrow\big(\text{x}^2+\frac{1}{\text{x}^2}\big)^2=6241$
$\Rightarrow \big(\text{x}^2\big)^2+2\big(\text{x}^2\big)\big(\frac{1}{\text{x}^2}\big)+\big(\frac{1}{\text{x}^2}\big)^2=6241$
$\Rightarrow \text{x}^4+2+\frac{1}{\text{x}^4}=6241$
$\Rightarrow \text{x}^4+\frac{1}{\text{x}^4}=6239$
View full question & answer→Question 135 Marks
If $\text{x}+\frac{1}{\text{x}}=12,$ find the value of $\text{x}−\frac{1}{\text{x}}.$
AnswerLet us consider the following equation: $\text{x}+\frac{1}{\text{x}}=12$
Squaring both sides, we get: $\big(\text{x}+\frac{1}{\text{x}}\big)^2=(12)^2=144$
$\Rightarrow\big(\text{x}+\frac{1}{\text{x}}\big)^2=144$
$\Rightarrow \text{x}^2+2\times \text{x}\times \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] $
$\Rightarrow \text{x}^2+2+\frac{1}{\text{x}^2}=144$
$\Rightarrow \text{x}^2+\frac{1}{\text{x}^2}=142$ (Subtracting $2$ from both sides)
Now $\big(\text{x}−\frac{1}{\text{x}}\big)^2=\text{x}^2−2\times \text{x}\times \frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2$
$=\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)
View full question & answer→Question 145 Marks
Write the following squares of binomials as trinomials: $\big(\frac{\text{x}}{\text{y}}−\frac{\text{y}}{\text{x}})^2$
AnswerWe 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.
$\big(\frac{\text{x}}{\text{y}}−\frac{\text{y}}{\text{x}})^2$
$=\big(\frac{\text{x}}{\text{y}})^2−2\big(\frac{\text{x}}{\text{y}}\big)\big(\frac{\text{y}}{\text{x}}\big)+\big(\frac{\text{y}}{\text{x}})^2$
$=\frac{\text{x}^2}{\text{y}^2}−2+\frac{\text{y}^2}{\text{x}^2}$
View full question & answer→Question 155 Marks
Multiply the monomial by the binomial and find the value of $x = -1, y = 0.25$ and $z = 0.05: 15y^2(2 - 3x)$
AnswerTo find the product, we will use distributive law as follows:
$ 15 y^2(2-3 x) $
$ =15 y^2 \times 2-15 y^2 \times 3 x $
$ =30 y^2-45 x y^2 $
Substituting $x = -1$ and $y = 0.25$ in the result, we get:
$30 y^2-45 x y^2$
$=30(0.25)^2-45(-1)(0.25)^2$
$= 30 × 0.0625 - [45 × (−1) × 0.0625] $
$= 30 × 0.0625 - [45 × (−1) × 0.0625] $
$= 1.875 - (-2.8125) $
$= 1.875 + 2.8125$
$ = 4.6875$
View full question & answer→Question 165 Marks
Find the value of $\left(5 x^6\right) \times\left(-1.5 x^2 y^3\right) \times\left(-12 x y^2\right)$ when $x=1, y=0.5$.
AnswerFirst multiply the expressions and then substitute the values for the variables. To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m× a^n= a^{m+n}$. We have:
$ \left(5 x^6\right) \times\left(-1.5 x^2 y^3\right) \times\left(-12 x y^2\right)$
$ =[5 \times(-1.5) \times(-12)] \times\left(x^6 \times x^2 \times x\right) \times\left(y^3 \times y^2\right) $
$ =[5 \times(-1.5) \times(-12)] \times\left(x^{6+2+1}\right) \times\left(y^{3+2}\right) $
$ =90 x^9 y^5 $
$ \therefore\left(5 x^6\right) \times\left(-1.5 x^2 y^3\right) \times\left(-12 x y^2\right)=90 x^9 y^5 $
Substituting $x = 1$ and $y = 0.5$ in the result, we get:
$=90 x^9 y^5$
$=90(1)^9(0.5)^5$
$=90 \times 1 \times 0.03125$
$=2.8125$
Thus, the answer is $2.8125$.
View full question & answer→Question 175 Marks
If $\text{x}+\frac{1}{\text{x}}=20,$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
AnswerLet us consider the following equation: $\text{x}+\frac{1}{\text{x}}=20,$
Squaring both sides, we get: $\big(\text{x}+\frac{1}{\text{x}}\big)^2=(20)^2=400$
$\Big[\big(\text{a}+\text{b}\big)^2=\text{a}^2 +\text{b}^2 +2\text{ab}\Big]$
$\big(\text{x}+\frac{1}{\text{x}}\big)^2=400$
$\Rightarrow \text{x}^2+2\times \text{x}\times \frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2=400 $
$\Rightarrow \text{x}^2+2+\frac{1}{\text{x}^2}=400$
$\Rightarrow \text{x}^2+\frac{1}{\text{x}^2}=398$ (Subtracting $2$ from both sides) Thus, the answer is $398$.
View full question & answer→Question 185 Marks
Find product: Express product as a monomials and verify the result in each case for $x = 1: (3x) × (4x) × (-5x)$
AnswerWe have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,
$a^m× a^n= a^{m+n}$.
We have:
$ (3 \mathrm{x}) \times(4 \mathrm{x}) \times(-5 \mathrm{x}) $
$ =(3 \times 4 \times(-5) \times(\mathrm{x} \times \mathrm{x} \times \mathrm{x}) $
$ =\left(3 \times 4 \times(-5) \times\left(\mathrm{x}^{1+1+1}\right)\right. $
$ =-60 \mathrm{x}^3$
Substituting $x=1$ in $LHS$, we get:
$ \text { LHS }=(3 x) \times(4 x) \times(-5 x) $
$ =(3 \times 1) \times(4 \times 1) \times(-5 \times 1) $
$ =-60$
Putting $x=1$ in $RHS$, we get:
$ \text { RHS }=-60 \times 3 $
$ =-60(1)^3 $
$ =-60 \times 1 $
$ =-60$
Sibce, $LHS = RHS$ for $x = 1$, therefore, the result is correct
Thus, the answer is $-60x^3$
View full question & answer→Question 195 Marks
Evaluate $\left(-8 x^2 y^6\right) \times(-20 x y)$ for $x=2.5$ and $y=1$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m× a^n= a^{m+n}$.
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 $
$ \left(-8 x^2 y^6\right) \times(-20 x y)=160 x^3 y^7$
Substituting $x=2.5$ and $y=1$ in the result, we get:
$ =160 x^3 y^7 $
$ =160(2.5)^3(1)^7 $
$ =160 \times 15.625 $
$ =2500$
Thus, the answer is $2500$.
View full question & answer→Question 205 Marks
Find the product $ -3 y\left(x y+y^2\right) $ and find its value for $x = 4$ and $y = 5.$
AnswerTo find the product, we will use distributive law as follows:
$ -3 y\left(x y+y^2\right) $
$ =-3 y \times x y+(-3 y) \times y^2 $
$ =-3 x y^{1+1}-3 y^{1+2} $
$ =-3 x y^2-3 y^3$
Substituting $x=4$ and $y=5$ in the result, we get:
$ -3 x y^2-3 y^3 $
$ =-3(4)(5)^2-3(5)^3 $
$ =-3(4)(25)-3(125) $
$ =-300-375 $
$ =-675$
Thus, the product is $\left(-3 x y^2-y^3-3 x y^2-3 y^3\right)$, and its value for $x=4$ and $y=5$ is $(-675)$.
View full question & answer→Question 215 Marks
Find the product:
$-\frac{8}{27}\text{xyz}\big(\frac{3}{2}\text{xyz}^2-\frac{9}{4}\text{xy}^2\text{z}^2\big)$
Answer To find the product, we will use distributive law as follows: $-\frac{8}{27}\text{xyz}\big(\frac{3}{2}\text{xyz}^2-\frac{9}{4}\text{xy}^2\text{z}^2\big)$
$=\Big\{\big(−\frac{8}{27}\text{xyz}\big)\big(\frac{3}{2}\text{xyz}^2\big)\Big\}−\Big\{\big(−\frac{8}{27}\text{xyz}\big)\big(\frac{9}{4}\text{xy}^2\text{z}^{3}\big)\Big\}$
$=\Big\{\big(−\frac{8}{27}\times\frac{3}{2}\big)\big(\text{x}×\text{x}\big)×\big(\text{y}×\text{y}\big)×\big(\text{z}×\text{z}^2\big)\Big\}\\−\Big\{\big(−\frac{8}{27}×\frac{9}{4}\big)\big(\text{x}×\text{x}\big)×\big(\text{y}×\text{y}^2)×\big(\text{z}×\text{z}^3\big) \Big\}$
$=\Big\{\big(−\frac{8}{27}\times\frac{3}{2}\big)\big(\text{x}^{1+1}\text{y}^{1+1}\text{z}^{1+2}\big)\Big\}\\−\Big\{\big(−\frac{8}{27}×\frac{9}{4}\big)\big(\text{x}^{1+1}\text{y}^{1+2}\text{z}^{1+3}\big)\Big\}$
$=−\frac{4}{9}\text{x}^2\text{y}^2\text{z}^3+\frac{2}{3}\text{x}^2\text{y}^3\text{z}^4$
Thus, the answer is $−\frac{4}{9}\text{x}^2\text{y}^2\text{z}^3+\frac{2}{3}\text{x}^2\text{y}^3\text{z}^4$
View full question & answer→Question 225 Marks
Take away: $\frac{6}{5}\text{x}^{2}-\frac{4}{5}\text{x}^3+\frac{5}{6}+\frac{3}{2}\text{x}$ from $\frac{\text{x}^3}{3}−\frac{5}{2}\text{x}^2+\frac{3}{5}\text{x}+\frac{1}{4}$
AnswerThe difference is given by: $\big(\frac{\text{x}^3}{3}−\frac{5}{2}\text{x}^2+\frac{3}{5}\text{x}+\frac{1}{4}\big)$
$-\big(\frac{6}{5}\text{x}^{2}-\frac{4}{5}\text{x}^3+\frac{5}{6}+\frac{3}{2}\text{x}\big)$
$\frac{\text{x}^3}{3}−\frac{5}{2}\text{x}^2+\frac{3}{5}\text{x}+\frac{1}{4}-\frac{6}{5}\text{x}^{2}$
$+\frac{4}{5}\text{x}^3-\frac{5}{6}-\frac{3}{2}\text{x}$
$=\frac{\text{x}^3}{3}+\frac{4}{5}\text{x}^3-\frac{5}{2}\text{x}^2-\frac{6}{5}\text{x}+\frac{3}{5}\text{x}$
$-\frac{3}{2}\text{x}+\frac{1}{4}-\frac{5}{6}$
$=\big(\frac{5+12}{15}\big)\text{x}^3+\big(\frac{−25−12}{10}\big)\text{x}^2$
$+\big(\frac{6−15}{10}\big)\text{x}+\big(\frac{6−20}{24}\big)$ (Collecting like terms)
$=\frac{17}{15}\text{x}^3-\frac{37}{10}\text{x}^2-\frac{9}{10}\text{x}-\frac{7}{12}$ (Combining like terms)
View full question & answer→Question 235 Marks
Find the following product and verify the result for $x = -1, y = -2: (3x - 5y) (x + y)$
AnswerTo multiply, we will use distributive law as follows:
$ (3 x-5 y)(x+y) $
$ =3 x(x+y)-5 y(x+y) $
$ =3 x^2+3 x y-5 x y-5 y^2 $
$ =3 x^2-2 x y-5 y^2 $
$ =\therefore(3 x-5 y)(x+y)=3 x^2-2 x y-5 y^2$
$\text { Now, we put } x=-1 \text { and } y=-2 \text { on both sides to verify the result. }$
$ \text { LHS }=(3 x-5 y)(x+y) $
$ =\{3(-1)-5(-2)\}\{-1+(-2)\} $
$ =(-3+10)(-3) $
$ =(7)(-3) $
$ =-21 $
$ \text { RHS }=3 x^2-2 x y-5 y^2 $
$ =3(-1)^2-2(-1)(-2)-5(-2)^2 $
$ =3 \times 1-4-5 \times 4 $
$ =3-4-20 $
$ =-21$
Because $LHS$ is equal to $RHS$, the result is verified.
View full question & answer→Question 245 Marks
Express product as a monomials and verify the result in each case for $x = 1: (5x^4) × (x^2)^3× (2x)^2$
AnswerWe have to find the product of the expression in order to express it as a monomial. To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^m× a^n= a^{m+n}$ and $(a^m)^n= a^{mn}$.
We have:
$ \left(5 x^4\right) \times\left(x^2\right)^3 \times(2 x)^2 $
$ =\left(5 x^4\right) \times\left(x^6\right) \times\left(2^2 \times x^2\right) $
$ =\left(5 \times 2^2\right) \times\left(x^4 \times x^6 \times x^2\right) $
$ =\left(5 \times 2^2\right) \times\left(x^{4+6+2}\right) $
$ =20 x^{12} $
$ \therefore\left(5 x^4\right) \times\left(x^2\right)^3 \times(2 x)^2=20 x^{12} $
$ \text { LHS }=\left(5 x^4\right) \times\left(x^2\right)^3 \times(2 x)^2 $
$ =\left(5 \times 1^4\right) \times\left(1^2\right)^3 \times(2 \times 1)^2 $
$ =(5 \times 1) \times\left(1^6\right) \times(2)^2 $
$ =5 \times 1 \times 4 $
$ =20$
$\text { Put } x=1 \text { in RHS, we get: }$
$ \text { RHS }=20 x^{12} $
$ =20 \times(1)^{12} $
$ =20 \times 1 $
$ =20$
$\because$ $LHS = RHS$ for $x = 1$, therefore, the result is correct.
Thus, the answer is $20x^{12}$.
View full question & answer→Question 255 Marks
Subtract the sum of $3l - 4m - 7n^2$ and $2l + 3m - 4n^2$ from the sum of $9l + 2m - 3n^2$ and $-3l + m + 4n^2$.
Answer We have to subtract the sum of $\left(3l-4 m-7 n^2\right)$ and $\left(2l+3 m-4 n^2\right)$ from the sum of $\left(9l+2 m-3 n^2\right)$ and $\left(-3l+m+4 n^2\right)$
$ \left\{\left(9l+2 m-3 n^2\right)+\left(-3l+m+4 n^2\right)\right\}-\left\{\left(3l-4 m-7 n^2\right)+\left(2l+3 m-4 n^2\right)\right. $
$ =\left(9l-3l+2 m+m-3 n^2+4 n^2\right)-\left(3l+2l-4 m+3 m-7 n^2-4 n^2\right) $
$ =\left(6l+3 m+n^2\right)-\left(5l-m-ll n^2\right) \text { (Combining like terms inside the parentheses) } $
$ =6l+3 m+n^2-5l+m+ll n^2 $
$ =6l-5l+3 m+m+n^2+ll n^2 \text { (Collecting like terms) } $
$ =l+4 m+l2 n^2 \text { (Combining like terms) }$
Thus, the required solution is $I+4 m+l2 n^2$.
View full question & answer→Question 265 Marks
Simplify: $ (2.5 p-1.5 q)^2-(1.5 p-2.5 q)^2 $
AnswerTo simplify, we will proceed as follows:
$ (2.5 p-1.5 q)^2-(1.5 p-2.5 q)^2 $
$=(2.5 p)^2+(1.5 q)^2-2(2.5 p)(1.5 q)-\left[(1.5 p)^2+(2.5 q)^2-2(1.5 p)(2.5 q)\right] $
$ =(2.5 p)^2-(1.5 p)^2+(1.5 q)^2-(2.5 q)^2 $
$ =[(2.5 p+1.5 p)(2.5 p-1.5 p)]+[(1.5 q+2.5)(1.5 q-2.5 q)] $
$ =4 p \times p+4 q \times(-q) $
$ =4 p^2-4 q^2 $
$ =4\left(p^2-4 q^2\right) $
View full question & answer→Question 275 Marks
Evaluate following when $x = 2, y = −1$. $$$\big(\frac{3}{5}\text{x}^2\text{y}\big)×\big(−\frac{15}{4}\text{xy}^2\big)×\big(\frac{7}{9}\text{x}^2\text{y}^2)$
AnswerTo 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:$\big(\frac{3}{5}\text{x}^2\text{y}\big)×\big(−\frac{15}{4}\text{xy}^2\big)×\big(\frac{7}{9}\text{x}^2\text{y}^2)$
$=\big[\frac{3}{5}×\big(−\frac{15}{4}\big)×\frac{7}{9}\big]×\big(\text{x}^2×\text{x}×\text{x}^2\big)\\×\big(\text{y}×\text{y}^2×\text{y}^2)$
$=\big[\frac{3}{5}×\big(−\frac{15}{4}\big)×\frac{7}{9}\big]×\big(\text{x}^{2+1+2}\big)\\×\big(\text{y}^{1+2+2}\big) $
$=−\frac{7}{4}\text{x}^5\text{y}^5$
$\therefore\big(\frac{3}{5}\text{x}^2\text{y}\big)×\big(−\frac{15}{4}\text{xy}^2\big)×\big(\frac{7}{9}\text{x}^2\text{y}^2)=−\frac{7}{4}\text{x}^5\text{y}^5$
Substituting $x = 2$ and $y = -1$ in the result, we get:$−\frac{7}{4}\text{x}^5\text{y}^5$
$=−\frac{7}{4}(2)^5(−1)^5$
$=\big(−\frac{7}{4}\big)×32×(−1)$
$=56$
Thus, the answer is $56$.
View full question & answer→Question 285 Marks
Evaluate $ \left(2.3 a^5 b^2\right) \times\left(1.2 a^2 b^2\right) $ when $a = 1$ and $b = 0.5$.
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m× a^n= a^{m+n}$.
$ \left(2.3 a^5 b^2\right) \times\left(1.2 a^2 b^2\right) $
$ =(2.3 \times 1.2) \times\left(a^5 \times a^2\right) \times\left(b^2 \times b^2\right) $
$ =(2.3 \times 1.2) \times\left(a^5+2\right) \times\left(b^2+2\right) $
$ =2.76 a^7 b^4 $
$ \left(2.3 a^5 b^2\right) \times\left(1.2 a^2 b^2\right)=2.76 a^7 b^4$
Substituting $a=1$ and $b=0.5$ in the result, we get
$ 2.76 a^7 b^4 $
$ =2.76(1)^7(0.5)^4 $
$ =2.76 \times 1 \times 0.0625 $
$ =0.1725$
Thus, the answer is $= 0.1725$
View full question & answer→Question 295 Marks
Express product as a monomials and verify the result in each case for $x = 1$: $\big(4\text{x}^2\big) × (-3\text{x}) × \big(\frac{4}{5}\text{x}^3\big)$
AnswerWe have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^m \times a^n=a^{m+n}$.
$\big(4\text{x}^2\big) × (-3\text{x}) × \big(\frac{4}{5}\text{x}^3\big)$
$=\big[4×(−3)×\frac{4}{5}\big]×\big(\text{x}^2×\text{x}×\text{x}^3\big)$
$=\big[4×(−3)×\frac{4}{5}\big]×\big(\text{x}^{2+1+3}\big)$
$=−\frac{48}{5}\text{x}^6$
$\therefore\big(4\text{x}^2\big) × (-3\text{x}) × \big(\frac{4}{5}\text{x}^3\big)=−\frac{48}{5}\text{x}^6$
Substituting $x = 1$ in $LHS$, we get:
$\text{LHS}=\big(4\text{x}^2\big) × (-3\text{x}) × \big(\frac{4}{5}\text{x}^3\big)$
$=\big(4×1^2\big)×(-3×1)×\big(\frac{4}{5}×1^3)$
$=4×(−3)×\frac{4}{5}$
$=−\frac{48}{5}$
Putting $x = 1$ in $RHS$, we get:
$\text{RHS}=−\frac{48}{5}\text{x}^6$
$=−\frac{48}{5}×1^6$
$=−\frac{48}{5}$
$\because$ $LHS = RHS$ for $x = 1$, therefore, the result is correct
Thus, the answer is $−\frac{48}{5}\text{x}^6.$
View full question & answer→Question 305 Marks
Simplify:
$ (2 x-1)(2 x+1)\left(4 x^2+1\right)\left(16 x^4+1\right) $
AnswerTo simplify, we will proceed as follows:
$ (2 x-1)(2 x+1)\left(4 x^2+1\right)\left(16 x^4+1\right) $
$ =\left((2 x)^2-1^2\right)\left(4 x^2+1\right)\left(16 x^4+1\right)\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =\left(4 x^2-1\right)\left(4 x^2+1\right)\left(16 x^4+1\right) $
$ =\left\{\left(4 x^2\right)^2-\left(1^2\right)^2\right\}\left(16 x^4+1\right)\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =\left(16 x^4-1\right)\left(16 x^4+1\right)$
$ =\left(16 x^4\right)^2-1^2\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =256 x^8-1 $
View full question & answer→Question 315 Marks
Simplify: $ (5 x+3)(x-1)(3 x-2) $
AnswerTo simplify, we will proceed as follows:
$ (5 x+3)(x-1)(3 x-2) $
$ =[(5 x+3)(x-1)](3 x-2) $
$ =[5 x(x-1)+3(x-1)](3 x-2) $
$ =\left[5 x^2-5 x+3 x-3\right](3 x-2) \text { (Distributive law) } $
$ =\left[5 x^2-2 x-3\right](3 x-2) $
$ =3 x\left(5 x^2-2 x-3\right)-2\left(5 x^2-2 x-3\right) $
$ =15 x^3-6 x^2-9 x-\left[10 x^2-4 x-6\right] $
$ =15 x^3-6 x^2-9 x-10 x^2+4 x+6 $
$ =15 x^3-6 x^2-10 x^2-9 x+4 x+6 \text { (Rearranging) } $
$ =15 x^3-16 x^2-5 x+6 \text { (Combining like terms) }$
Thus, the answer is $15 x^3-16 x^2-5 x+6$
View full question & answer→Question 325 Marks
Multiply the monomial by the binomial and find the value of$ x = -1, y = 0.25$ and $z = 0.05$:
$ x z\left(x^2+y^2\right) $
AnswerTo find the product, we will use distributive law as follows:
$ x z\left(x^2+y^2\right) $
$ =x z \times x^2+x z \times y^2 $
$ =x^3 z+x y^2 z $
Substituting $x = -1, y = 0.25$ and $z = 0.05$ in the result, we get:
$ x^3 z+x y^2 z $
$= (-1)^3(0.05) + (-1)(0.25)^2(0.05)$
$= (-1)(0.05) + (-1)(0.0625)(0.05)$
$= -0.05 - 0.003125$
$= -0.053125$
View full question & answer→Question 335 Marks
Simplify:
$ \left(2 x^2+3 x-5\right)\left(3 x^2-5 x+4\right) $
AnswerTo simplify, we will proceed as follows:
$ \left(2 x^2+3 x-5\right)\left(3 x^2-5 x+4\right) $
$ =2 x^2\left(3 x^2-5 x+4\right)+3 x\left(3 x^2-5 x+4\right)-5\left(3 x^2-5 x+4\right) \text { (Distributive law) } $
$ =6 x^4-10 x^3+8 x^2+9 x^3-15 x^2+12 x-15 x^2+25 x-20 $
$ =6 x^4-10 x^3+9 x^3+8 x^2-15 x^2-15 x^2+12 x+25 x-20 \text { (Rearranging) } $
$ =6 x^4-x 3-22 x^2+36 x-20$
Thus, the answer is $6 x^4-x 3-22 x^2+36 x-20$.
View full question & answer→Question 345 Marks
Simplify:
$ \left(x^2-3 x+2\right)(5 x-2)-\left(3 x^2+4 x-5\right)(2 x-1) $
AnswerTo simplify, we will proceed as follows:
$ \left(x^2-3 x+2\right)(5 x-2)-\left(3 x^2+4 x-5\right)(2 x-1) $
$ =\left[\left(x^2-3 x+2\right)(5 x-2)\right]-\left[\left(3 x^2+4 x-5\right)(2 x-1)\right] $
$ =\left[5 x\left(x^2-3 x+2\right)-2\left(x^2-3 x+2\right)\right]-\left[2 x\left(3 x^2+4 x-5\right)-1 \times\left(3 x^2+4 x-5\right)\right] \text { (Distributive law) } $
$ =\left[5 x^3-15 x^2+10 x-\left(2 x^2-6 x+4\right)\right]-\left[6 x^3+8 x^2-10 x-3 x^2-4 x+5\right] $
$ =5 x^3-15 x^2+10 x-2 x^2+6 x-4-6 x^3-8 x^2+10 x+3 x^2+4 x-5 $
$ =5 x^3-6 x^3-15 x^2-2 x^2-8 x^2+3 x^2+10 x+6 x+10 x+4 x-5-4 \text { (Rearranging) } $
$ =-x 3-22 x^2+30 x-9 \text { (Combining like terms) }$
Thus, the answer is $-x 3-22 x^2+30 x-9$.
View full question & answer→Question 355 Marks
Express product as a monomials and verify the result in each case for $ x = 1: \left(x^2\right)^3 \times(2 x) \times(-4 x) \times(5) $
AnswerWe have to find the product of the expression in order to express it as a monomial. To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^m× a^n= a^{m+n}$ and $(a^m)^n= a^{mn}$.
We have:
$ \left(x^2\right)^3 \times(2 x) \times(-4 x) \times(5) $
$ =\left(x^6\right) \times(2 x) \times(-4 x) \times 5 $
$ =[2 \times(-4) \times 5] \times\left(x^6 \times x \times x\right) $
$ =[2 \times(-4) \times 5] \times\left(x^{6+1+1}\right) $
$ =-40 x^8 $
$ \therefore\left(x^2\right)^3 \times(2 x) \times(-4 x) \times(5)=-40 x^8 $
$ \text { LHS }=\left(x^2\right)^3 \times(2 x) \times(-4 x) \times(5) $
$ =\left(1^2\right)^3 \times(2 \times 1) \times(-4 \times 1) \times 5 $
$ =1^6 \times 2 \times(-4) \times 5 $
$ =1 \times 2 \times(-4) \times 5 $
$ =-40$
Putting $x=1$ in $RHS$, we get:
$ \text { RHS }=-40 x^8 $
$ =-40(1)^8 $
$ =-40 \times 1 $
$ =-40$
$\because$ $LHS = RHS$ for $x = 1$; therefore, the result is correct Thus, the answer is $-40x^8$.
View full question & answer→Question 365 Marks
Simplify: $(5 - x)(6 - 5x)( 2 - x)$
AnswerTo simplify, we will proceed as follows:
$ (5-x)(6-5 x)(2-x) $
$ =[(5-x)(6-5 x)](2-x) $
$ =[5(6-5 x)-x(6-5 x)](2-x) \text { (Distributive law) } $
$ =\left(30-25 x-6 x+5 x^2\right)(2-x) $
$ =\left(30-31 x+5 x^2(2-x)\right. $
$ =2\left(30-31 x+5 x^2\right)-x\left(30-31 x+5 x^2\right) $
$ =60-62 x+10 x^2-30 x+31 x^2-5 x^3 $
$ =60-62 x+10 x^2-30 x+31 x^2-5 x^3 $
$ =60-62 x-30 x+10 x^2+31 x^2-5 x^3 \text { (Rearranging) } $
$ =60-92 x+41 x^2-5 x^3 \text { Combining like terms) }$
Thus, the answer is $60-92 x+41 x^2-5 x^3$.
View full question & answer→Question 375 Marks
Simplify: $(3x - 2)(2x - 3) + (5x - 3)(x + 1)$
AnswerTo simplify, we will proceed as follows:
$ (3 x-2)(2 x-3)+(5 x-3)(x+1) $
$ =[(3 x-2)(2 x-3)]+[(5 x-3)(x+1)] $
$ =[3 x(2 x-3)-2(2 x-3)]+[5 x(x+1)-3(x+1)] \text { (Distributive law) } $
$ =6 x^2-9 x-4 x+6+5 x^2+5 x-3 x-3 $
$ =6 x^2+5 x^2-9 x-4 x+5 x-3 x-3+6 $
$ =11 x^2-11 x+3 \text { (Combining like terms) }$
Thus, the answer is $11 x^2-11 x+3$.
View full question & answer→Question 385 Marks
Simplify: $(3x + 2y)(4x + 3y) - (2x - y)(7x - 3y)$
AnswerTo simplify, we will proceed as follows:
$ (3 x+2 y)(4 x+3 y)-(2 x-y)(7 x-3 y) $
$ =[(3 x+2 y)(4 x+3 y)]-[(2 x-y)(7 x-3 y)] $
$ =[3 x(4 x+3 y)+2 y(4 x+3 y)]-[2 x(7 x-3 y)-y(7 x-3 y)] \text { (Distributive law) } $
$ =12 x^2+9 x y+8 x y+6 y^2-\left[14 x^2-6 x y-7 x y+3 y^2\right] $
$ =12 x^2+9 x y+8 x y+6 y^2-14 x^2+6 x y+7 x y-3 y^2 $
$ =12 x^2-14 x^2+9 x y+8 x y+6 x y+7 x y+6 y^2-3 y^2 $
$ =-2 x^2+30 x y+3 y^2 \text { (Rearranging) }$
Thus, the answer is $-2 x^2+30 x y+3 y^2$.
View full question & answer→Question 395 Marks
Evaluate $\left(3.2 x^6 y^3\right) \times\left(2.1 x^2 y^2\right)$ when $x = 1$ and $y = 0.5$.
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m× a^n= a^{m+n}$.
We have:
$ \left(3.2 x^6 y^3\right) \times\left(2.1 x^2 y^2\right) $
$ =(3.2 \times 2.1) \times\left(x^6 \times x^2\right) \times\left(y^3 \times y^2\right) $
$ =(3.2 \times 2.1) \times\left(x^{6+2}\right) \times\left(y^{3+2}\right) $
$ =6.72 x^8 y^5$
Substituting $\mathrm{x}=1$ and $\mathrm{y}=0.5$ in the result, we get:
$ 6.72 x^8 y^5 $
$ =6.72(1)^8(0.5)^5 $
$ =6.72 \times 1 \times 0.03125 $
$ =0.21$
Thus, the answer is $0.21$.
View full question & answer→Question 405 Marks
Find the product $ 24 x^2(1-2 x) $ and evaluate its value for $x = 3$.
AnswerTo find the product, we will use distributive law as follows:
$ 24 x^2(1-2 x) $
$ =24 x^2 \times 1-24 x^2 \times 2 x $
$ =24 x^2-48 x^{1+2} $
$ =24 x^2-48 x^3 $
Substituting $x=3$ in the result, we get:
$ 24 x^2-48 x^3 $
$ =24(3)^2-48(3)^3 $
$ =24 \times 9-48 \times 27 $
$ =216-1296 $
$ =-1080 $
Thus, the product is $\left(24 x^2-48 x^3\right)$ and its value for $x=3$ is $-1080$ .
View full question & answer→Question 415 Marks
Multiply the monomial by the binomial and find the value of $x = -1, y = 0.25$ and $ z = 0.05: -3 x\left(y^2+z^2\right) $
AnswerTo find the product, we will use distributive law as follows:
$ -3 x\left(y^2+z^2\right) $
$ =-3 x \times y^2+(-3 x) \times z^2 $
$ =-3 x y^2-3 x z^2$
Substituting $x=-1, y=0.25$ and $z=0.05$ in the result.
we get:
$ -3 x y^2-3 x z^2 $
$ =-3(-1)(0.25)^2-3(-1)(0.05)^2 $
$ =-3(-1)(0.0625)-3(-1)(0.0025) $
$ =01875+0.0075 $
$ =0.195$
View full question & answer→Question 425 Marks
Simplify the following using the identities: $\frac{8.63×8.63−1.37\times1.37}{0.726}$
AnswerLet us consider the following expression:
$\frac{8.63\times 8.63−1.37\times1.37}{0.726}=\frac{8.63^2−1.37^2}{0.726}$
Using the identity $(a + b) (a - b) = a^2- b^2$
$\frac{8.63\times 8.63−1.37\times1.37}{0.726}=\frac{8.63^2−1.37^2}{0.726}$
$=\frac{(8.63+1.37)(8.63−1.37)}{0.726}$
$\frac{8.63\times 8.63−1.37\times1.37}{0.726}=\frac{(8.63+1.37)(8.63−1.37)}{0.726}$
$\frac{8.63\times 8.63−1.37\times1.37}{0.726}=\frac{10\times 7.26}{0.726}$
$\frac{8.63\times 8.63−1.37\times 1.37}{0.726}=100$
Thus, the answer is $100$.
View full question & answer→Question 435 Marks
Simplify: $(5x - 3)(x + 2) - (2x + 5)(4x - 3)$
AnswerTo simplify, we will proceed as follows:
$ (5 x-3)(x+2)-(2 x+5)(4 x-3)$
$ =[(5 x-3)(x+2)]-[(2 x+5)(4 x-3)] $
$ =[5 x(x+2)-3(x+2)]-[2 x(4 x-3)+5(4 x-3)] \text { (Distributive law) }$
$ =5 x^2+10 x-3 x-6-8 x^2+6 x-20 x+15 $
$ =5 x^2-8 x^2+10 x-3 x+6 x-20 x-6+15 \text { (Rearranging) } $
$ =-3 x^2-7 x+9 $
View full question & answer→Question 445 Marks
Simplify:
$ x^2(x-y) y^2(x+2 y) $
AnswerTo simplify, we will proceed as follows:
$ x^2(x-y) y^2(x+2 y) $
$ =\left[x^2(x-y)\right]\left[y^2(x+2 y)\right] $
$ =\left(x^3-x^2 y\right)\left(x y^2+2 y^3\right) $
$ =x^3\left(x y^2+2 y^3\right)-x^2 y\left(x y^2+2 y^3\right) $
$ =x^4 y^2+2 x^3 y^3-\left[x^3 y^3+2 x^2 y^4\right] $
$ =x^4 y^2+2 x^3 y^3-x^3 y^3-2 x^2 y^4 $
$ =x^4 y^2+x^3 y^3-2 x^2 y^4$
Thus, the answer is $x^4 y^2+x^3 y^3-2 x^2 y^4$.
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