Tag: solving linear equations

Questions Related to solving linear equations

If $\dfrac {2}{3x + 12} = \dfrac {2}{3}$, then the value of $x + 4 $ is

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

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

  2. $1$

  3. $\dfrac {3}{2}$

  4. $2$

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Correct Option: B
Explanation:

Given $\dfrac { 2 }{ 3x+12 } =\dfrac { 2 }{ 3 } $

$\Rightarrow 3(2)=2(3x+12)$
$\Rightarrow 6=6x+24$
$\Rightarrow 6x=-18$
$\Rightarrow x=-3$
Therefore $x+4=-3+4=1$

If $\sqrt[5]{\cfrac{g-1}{4}} = \cfrac{1}{3}$, then find the value of $g$.

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

  1. $0.984$

  2. $0.996$

  3. $1.004$

  4. $1.016$

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Correct Option: D
Explanation:

Given, $\sqrt [ 5 ]{ \dfrac { g-1 }{ 4 }  } =\dfrac { 1 }{ 3 } $

$ \Rightarrow  $ $\cfrac{g-1}{4} = \cfrac { 1 }{ { 3 }^{ 5 } } =\cfrac { 1 }{ 243 } $
$ \Rightarrow  $ $ g-1 = \cfrac{4}{243}$
$ \Rightarrow  $ $ g = 1 + \cfrac{4}{243} = 1.016$

The area of square $ABCD$ is three-fourths the area of parallelogram $EFGH$. The area of parallelogram $EFGH$ is one-third the area of trapezoid $IJKL$. If square $ABCD$ has an area of $125$ square feet, calculate the area of trapezoid $IJKL$, in square feet.

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

  1. $75$

  2. $225$

  3. $350$

  4. $500$

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Correct Option: D
Explanation:

Given, area of square $ABCD$ is three fourth of area of parallelogram $EFGH$,

And the area of parallelogram $EFGH$ is one-third of the area of trapezoid $IJKL$ and area of square $ABCD$ is $125$.
Let the area of trapezoid $IJKL$ is $x$
Then  area of  parallelogram $EFGH =$ $\dfrac{1}{3}x$
And  area of square $ABCD=$ $\dfrac{3}{4}$ area of  parallelogram $EFGH=$ $\dfrac{3}{4}\times \dfrac{1}{3}x=\dfrac{1}{4}x$
But area of square $ABCD =125$
$\therefore \dfrac{1}{4}x=125$
$\Rightarrow x=500$
So, area of trapezoid $IJKL=500$.

Find the value of $x: \dfrac {1}{x} + \dfrac {4}{5x} = \dfrac {2}{x + 5}$

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

  1. $0.71$

  2. $3.57$

  3. $5.8$

  4. $45$

Show answer
Correct Option: D
Explanation:
Given  $\dfrac { 2 }{ x+5 } =\dfrac { 1 }{ x } +\dfrac { 4 }{ 5x } $
Taking RHS:

$\dfrac { 1 }{ x } +\dfrac { 4 }{ 5x } $

LCM is $5x$
$\Rightarrow \dfrac { 5 }{ 5x } +\dfrac { 4 }{ 5x } \\ \Rightarrow \dfrac { 9 }{ 5x } $
Now taking LHS:
$\dfrac { 2 }{ x+5 } $
LHS $=$ RHS
$\dfrac { 2 }{ x+5 } =\dfrac { 9 }{ 5x } $
$\Rightarrow 5x\times 2=9(x+5)\\ \Rightarrow 10x=9x+45\\ \Rightarrow x=45$

Find the value of $\dfrac {4}{y} + 4$ given that $\dfrac {4}{y} + 4 = \dfrac {20}{y} + 20$

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

  1. $-1$

  2. $0$

  3. $1$

  4. $4$

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Correct Option: B
Explanation:
The value of $\frac{20}{y}+20=5\left ( \frac{4}{y}+4 \right )$
Given $\frac{4}{y}+4=5\left ( \frac{4}{y}+4 \right )$
This when possible then value of $\frac{4}{y}+4=0$

Compute the approximate value of $x$: $\sqrt [3]{\dfrac {2x + 3}{5}} = \dfrac {2}{3}$

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

  1. $-0.76$

  2. $-0.69$

  3. $-0.67$

  4. $0.69$

  5. $0.76$

Show answer
Correct Option: A
Explanation:
Given is $\sqrt [ 3 ]{ \dfrac { 2x+3 }{ 5 }  } = \dfrac { 2 }{ 3 } $
Now raising the power to $3$ on both sides, we get
$\dfrac { 2x+3 }{ 5 } = { (2/3) }^{ 3 }\\ \Rightarrow \dfrac { 2x+3 }{ 5 } =\dfrac { 8 }{ 27 } \\ \Rightarrow 54x+81=40\\ \Rightarrow x=-0.76$

If $\dfrac{3}{9}=\dfrac{3}{x+2}$, what is the value of $x$?

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

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

  2. $\dfrac{7}{3}$

  3. $3$

  4. $7$

  5. $\dfrac{25}{3}$

Show answer
Correct Option: D
Explanation:

Given, $\dfrac {3}{9}=\dfrac {3}{x+2}$

On cross multiplying, we get
$\Rightarrow 3(x+2)=9(3)$
$\Rightarrow x=9-2$
$\Rightarrow x=7$

The square roots of Radhas and Krishs ages have a sum of $7$ and a difference of $1$. If Radha is older than Krish, how old is Radha?

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

  1. $13$

  2. $4$

  3. $9$

  4. $16$

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Correct Option: D
Explanation:

Let the ages of Radha and krish be $x^{2}$ and $y^{2}$ respectively

Then square roots of their age will be $x$ and $y$ respectively.

Then according to the question,

$x+y=7$ ..(1)

$x-y=1$...(2)

Adding (1) and (2)

$2x=8$

$x=4$

From (1)

$4+y=7$

$y=3$

Therefore $x^{2}=16$ and $y^{2}=9$

As radha is older,hence age of radha is 16years.

What is the solution of $\displaystyle \frac{x-5}{2} - \frac{x-3}{5} = \frac{1}{2}$?

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

  1. $x = 5$

  2. $x = 7$

  3. $x = 8$

  4. $x = 9$

Show answer
Correct Option: C
Explanation:

Given ,$\dfrac{x-5}{2}-\dfrac{x-3}{5}=\dfrac{1}{2}$

$\dfrac{5x-25-2x+6}{5\times 2}=\dfrac{1}{2}$

$3x-19=5$

$3x=24$

$x=8$

If $x=\displaystyle\frac{1}{\displaystyle 2-\frac{1}{\displaystyle 2-\frac{1}{2-x}}}, (x\neq 2)$, then the value of x is ________?

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

  1. $1$

  2. $3$

  3. $2$

  4. $5$

Show answer
Correct Option: A
Explanation:

$x=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}$


$\Rightarrow x=\dfrac{1}{2-\dfrac{2-x}{3-2x}}$


$\Rightarrow x=\dfrac{3-2x}{4-3x}$

On solving, we get $x=1$
So, Option (A)