2 Marks Questions

Question 12 Marks

Solve: $\frac{3\text{x}}{5\text{x}+2}=-4$

Answer

$\frac{3\text{x}}{5\text{x}+2}=-4$ $\Rightarrow\frac{3\text{x}}{5\text{x}+2}=\frac{-4}{1}$ $\Rightarrow1(3\text{x})=-4(5\text{x}+2)$ (by cross multiplication) $\Rightarrow3\text{x}=-20\text{x}-8$ $\Rightarrow3\text{x}+20\text{x}=-8$ $\Rightarrow23\text{x}=-8$ $\Rightarrow\text{x}=\frac{-8}{23}$ $\therefore\text{x}=\frac{-8}{23}$

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Question 22 Marks

If $\Big(\text{x}-\frac{1}{\text{x}}\Big)=4$ find the value of: $\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)$

Answer

From the first part: $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=18$ Squaring both the sides: $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=18^2$ $\Rightarrow(\text{x}^2)^2+2\times(\text{x}^2)\times\Big(\frac{1}{\text{x}^2}\Big)+\Big(\frac{1}{\text{x}^2}\Big)^2=324$ $\Rightarrow\text{x}^4+2+\frac{1}{\text{x}^4}=324$ $\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=324-2$ $\therefore\text{x}^4+\frac{1}{\text{x}^4}=322$

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Question 32 Marks

Factorise:
$1-6 x+9 x^2$

Answer

$1-6 x+9 x^2$
$=9 x^2-6 x+1$
$=9 x^2-3 x-3 x+1$
$=3 x(3 x-1)-1(3 x-1)$
$=(3 x-1)(3 x-1)$
$=(3 x-1)^2$

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Question 42 Marks

Solve: $\frac{2\text{x}+7}{3\text{x}+5}=\frac{15}{17}$

Answer

$\frac{2\text{x}+7}{3\text{x}+5}=\frac{15}{17}$ $\Rightarrow17 ( 2\text{x} + 7 ) = 15 ( 3\text{x} + 5 ) $ $\Rightarrow34\text{x} + 119 = 45\text{x} + 75$ $\Rightarrow119 - 75 = 45\text{x} - 34\text{x}$ $\Rightarrow 44 = 11\text{x}$ $\Rightarrow \text{x} = \frac{44}{11}=4$ $\therefore\text{x}=4$

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Question 52 Marks

Factorise:
$63x^2y^2- 7$

Answer

$= 7(9x^2y^2- 1)$
$= 7((3xy)^2- (1)^2)$
[according to the formula $a^2- b^2= (a + b)(a - b)]$
$= 7(3xy + 1)(3xy - 1)$

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Question 62 Marks

Factorise:
$7 x^2-19 x-6$

Answer

$7 x^2-19 x-6$
$=7 x^2-21 x+2 x-6$
$=7 x(x-3)+2(x-3)$
$=(7 x+2)(x-3)$

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Question 72 Marks

Solve: $5x + 7 = 2x - 8$

Answer

$5\text{x} + 7 = 2\text{x}-8$
$\Rightarrow5\text{x}-\text{2x}=-8-7$
$\Rightarrow3\text{x}=-15$
$\Rightarrow\text{x}=\frac{-15}{3}=-5$
$\therefore\text{x}=-5$

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Question 82 Marks

Find the following products:
$(4a + 5b) × (5a - 6b)$

Answer

$(4a + 5b) × (5a - 6b) $
$=20 a^2-24 a b+25 a b-30 b^2$
$=20 a^2+a b-30 b^2$

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Question 92 Marks

Solve: $2z - 1 = 14 - z$

Answer

$2\text{z} - 1 = 14 - \text{z}$
$\Rightarrow2\text{z}+\text{z}=14+1$
$\Rightarrow3\text{z}=15$
$\Rightarrow\text{z}=\frac{15}{3}=5$
$\therefore\text{z}=5$

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Question 102 Marks

Solve: $\frac{\text{x}-5}{2}-\frac{\text{x}-3}{5}=\frac{1}{2}$

Answer

$\frac{\text{x}-5}{2}-\frac{\text{x}-3}{5}=\frac{1}{2}$
$\Rightarrow10\Big(\frac{\text{x}-5}{2}\Big)-10\Big(\frac{\text{x}-3}{5}\Big)=10\Big(\frac{1}{2}\Big)$
$($multiplying throughout by $10,$ which is the $L.C.M.$ of $2, 2$ and $5):$
$\Rightarrow5(\text{x}-5)-2(\text{x}-3)=5$
$\Rightarrow5\text{x}-2$

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Question 112 Marks

Two numbers are such that the ratio between them is $3 : 5.$ If each is increased by $10,$ the ratio between the new numbers so formed is $5 : 7.$ Find the original numbers.

Answer

Let us consider $x$ as the common multiple of both the number. Then, first number $= 3x$ Second number $= 5x$
$\therefore\frac{3\text{x}+10}{5\text{x}+10}=\frac{5}{7}$
$\Rightarrow7(3\text{x}+10)=5(5\text{x}+10)$ (by cross multiplication)
$\Rightarrow 21\text{x} + 70 = 25\text{x} + 50$
$\Rightarrow 21\text{x} - 25\text{x} = 50 - 70$
$\Rightarrow -4\text{x} = -20$
$\Rightarrow \text{x} =\frac{ -20}{-4} = 5$
Therefore, the common multiple of both the numbers is $5.$
First number $= 3x = 3 \times 5 = 15$
Second number $= 5x = 5 \times 5 = 25$

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Question 122 Marks

Find the following products:
$\left(6 x^2-x+8\right) \times\left(x^2-3\right)$

Answer

$\left(6 x^2-x+8\right) \times\left(x^2-3\right)$
$=6 x^4-18 x^2-x^3+3 x+8 x^2-24$
$=6 x^4-x^3-18 x^2+8 x^2+3 x-24$
$=6 x^4-x^3-10 x^2+3 x-24$

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Question 132 Marks

Solve: $8x + 3 = 27 + 2x$

Answer

$8\text{x} + 3 = 27 + 2\text{x}$
$\Rightarrow8\text{x}-\text{2x}=27-3$
$\Rightarrow6\text{x}=24$
$\Rightarrow\text{x}=\frac{24}{6}=4$
$\therefore\text{x}=4$

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