Tag: reducing simple equations to simpler form

Questions Related to reducing simple equations to simpler form

Solve for $x$ : $\displaystyle \sqrt{\frac{x\, -\,2}{x\, +\, 1}}\, =\, \frac{1}{2}$

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. 3

  2. 2

  3. 1

  4. 0

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Correct Option: A
Explanation:
$\cfrac { \sqrt { x-2 }  }{ \sqrt { x+1 }  }  = \cfrac { 1 }{ 2 }$
Squaring on both sides we get
$\cfrac { x-2 }{ x+1 }  = \cfrac { 1 }{ 4 } \\ 4(x-2) = (x+1)\\ 4x - 8 = x +1\\ 3x = 9\\ x = 3$

Solve for $x$ : $\displaystyle \frac{4}{3\sqrt{x}}\, =\, \frac{1}{2}$

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. ${5}\dfrac{6}{9}$

  2. ${7}\dfrac{1}{9}$

  3. $14$

  4. $27$

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Correct Option: B
Explanation:
$\cfrac{4}{3\sqrt{x}}=\cfrac{1}{2}$
Squre both the sides we get
$\cfrac{16}{9x}=\cfrac{1}{4}$
$\Rightarrow 9x=16\times 4$
$\Rightarrow 9x=64 \Rightarrow x=\cfrac{64}{9}$
$\Rightarrow x=7\cfrac{1}{9}$

Solve:
$x\, +\, y\, =\, 7xy$
$2x\, -\, 3y\, =\, -xy$

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $x\, =\, \displaystyle \frac{1}{8}$ and $y\, =\, \displaystyle \frac{1}{4}$

  2. $x\, =\, \displaystyle \frac{1}{7}$ and $y\, =\, \displaystyle \frac{1}{4}$

  3. $x\, =\, \displaystyle \frac{1}{2}$ and $y\, =\, \displaystyle \frac{1}{4}$

  4. $x\, =\, \displaystyle \frac{1}{3}$ and $y\, =\, \displaystyle \frac{1}{4}$

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Correct Option: D
Explanation:

Given equations are 

$x+y=7xy$ ....(1)
and $2x-3y=-xy$ ....(2)
Multiply equation (1) by $3$, we get
$3x+3y=21xy$ ....(3)
Add equations (2) and (3),
$5x=20xy$
$\Rightarrow y=\dfrac {1}{4}$
Put this value in equation (1), we get
$x+\dfrac {1}{4}=7x\times \dfrac {1}{4}$
$\Rightarrow x-\dfrac {7x}{4}=-\dfrac {1}{4}$
$\Rightarrow \dfrac {3x}{4}=\dfrac {1}{4}$
$\Rightarrow x=\dfrac {1}{3}$
Therefore, solution is $x=\dfrac {1}{3}, y=\dfrac {1}{4}$.

Find the value of $a$, if $x = 0.5$ is a solution of equation $ax^{2}\, +\, (a\, -\, 1)\,
x\, +\, 3\, =\, a$.

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $24$

  2. $15$

  3. $10$

  4. $8$

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Correct Option: C
Explanation:

Given equation is $ax^2+(a-1)x+3=a$
The equation can be written as
$ax^2+(a-1)x+3-a=0$
Since $x=0.5$ is a solution of the equation, 
$a(0.5)^2 + (a-1)(0.5) + 3 - a = 0$
$\Rightarrow 0.25a + 0.5a - 0.5 + 3 - a = 0$
$\Rightarrow   (0.25 + 0.5 -1)a - 0.5+3 = 0$
$\Rightarrow  -0.25a + 2.5 = 0$
$\Rightarrow  0.25a = 2.5$
$\Rightarrow  a = \dfrac{2.5}{0.25}$
$\Rightarrow a = 10$

The solution of the equation $\displaystyle \frac{2x+4}{3x-1}=\frac{4}{3}$ is 

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. 6

  2. 4

  3. $\displaystyle \frac{8}{3}$

  4. $\displaystyle \frac{3}{4}$

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Correct Option: C
Explanation:

Given equation is $\displaystyle \frac{2x+4}{3x-1}=\frac{4}{3}$
Cross multiplying, we get
$3(2x+4)=4(3x-1)$
$\Rightarrow 6x+12=12x-4\Rightarrow 6x=16$
$\Rightarrow \displaystyle x=\frac{16}{6}=\frac{8}{3}$.
Hence, the solution is $x=\cfrac{8}{3}$.

Find the value of $y$ in the equation : 
$\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $18$

  2. $9$

  3. $-38$

  4. $-32$

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Correct Option: C
Explanation:

Given, $\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$
$\Rightarrow \displaystyle \frac{2+y}{y-7}=\frac{4}{5}$

$\Rightarrow 5(2+y)=4(y-7) $ ....( cross multiplying )
$\Rightarrow 10+5y=4y-28$
$\Rightarrow y=-38$
Hence,. the solution is $x=-38$

A combination of locks requires 3 numbers to open. The second number is $\displaystyle 2d + 5$ greater than the first number. The third number is $\displaystyle 3d - 20$ less than the second number. The sum of the three numbers is $\displaystyle 10d + 9$. The first number is 

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $\displaystyle 5d-11$

  2. $\displaystyle 3d-7$

  3. $\displaystyle 2d+19$

  4. $\displaystyle 3d-11$

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Correct Option: B
Explanation:

Let the first number be $x$
Then, second number = $x + 2d + 5$
Third number = $x + 2d + 5 -(3d -20)$ = $x -d + 25$
Sum of the three numbers = $ x + x+2d + 5 + x - d +25$ = $ 3x + d +30$

Thus, $ 3x + d +30$ = $10d + 9$
$3x = 10d + 9 - d - 30$ = $9d - 21$
$x = 3d - 7$

If $\displaystyle \sqrt{\left ( x-1 \right )\left ( y+2 \right )}=7$, $x$ and $y$ being positive whole numbers, then the values of $x$ and $y$ are, respectively

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $8,5$

  2. $15,12$

  3. $22,19$

  4. $6,8$

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Correct Option: A
Explanation:

$ \sqrt{(x-1)(y+2)}=7\Rightarrow (x-1)(y+2)=7^{2}$
$ \Rightarrow (x-1)=7:and:(y+2)=7$
$ x=8$ and $\displaystyle y=5$

If $\displaystyle 4=\sqrt{x+\sqrt{x+\sqrt{x+....,}}}$ then the value of x will be 

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. 20

  2. 16

  3. 12

  4. 8

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Correct Option: C
Explanation:

$\displaystyle 4 = \sqrt{x+\sqrt{x+\sqrt{x+....}}}$
$\displaystyle \Rightarrow 4=\sqrt{x+4}$
$\displaystyle \Rightarrow 16={x+4}$
$\displaystyle \Rightarrow 16=x+4\Rightarrow x=12.$

$\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+...}}}$ equals 

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $\displaystyle 6^{\frac{2}{3}}$

  2. 6

  3. $\displaystyle 6^{\frac{1}{3}}$

  4. 3

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Correct Option: D
Explanation:

Let  $ x =\sqrt{6+\sqrt{6+\sqrt{6+...}}}$


sqare it on both sides


$x^2=\sqrt{6+\sqrt{6+\sqrt{6+...}}} = 6+x$


$\Rightarrow x^2=6+x $


$\Rightarrow x^2-x-6=0 $


$\Rightarrow x^2-3x+2x-6=0 $


$\Rightarrow x(x-3)+2(x-3)=0 $


$\Rightarrow (x-3)(x+2)=0 $


$either  x-3=0---> x=3 $


$ or  x+2=0 ----> x=-2 $


since result of sqrt of anything will be positive only, therefore,


Answer $x=3$