Free Linear Equations in One Variable 03 Practice Test - 8th Grade 

$\frac{x+5}{5x+8}=\frac{11}{38}$

38x + 190 = 55x + 88

55x -38x = 190 - 88

17x = 102

$x=\frac{102}{17}=6$

$\frac{5}{3x}$ = $\frac{25}{27}$

By cross multiplication, we get 

$5\times 27=3x\times 25$

$\Rightarrow 3x=\frac{5\times 27}{25}$

$\Rightarrow x=\frac{5\times 27}{3\times 25}$

$\Rightarrow x=\frac{9}{5}$

$=1.8$, which lies between 1 and 2.

Let the age of Sunil be $3x$.
Then, the age of Sunil's father would be $7x$.

Four years later,
Sunil's age = $3x+4$ and
Sunil's father's age = $7x+4$.

Then, the sum of their ages four years later $=3x+4+7x+4=10x+8$.

Given that, $10x+8=78$.
                     $\Rightarrow 10x=70\Rightarrow x=7$

$\therefore$ Present age of Sunil = 3 × 7
                                      = 21 years
Present age of Sunil's father = 7 × 7
                                               = 49 years 

$\frac{x+9}{4}+\frac{2x+3}{5}=10$ is the given equation.

We can check if $x=11$ is a solution by substituting this value in the LHS to check whether it is equal to RHS or not 

$\frac{(11+9)}{4}+\frac{(22+3)}{5}$
= 5 + 5
= 10

Thus we see that $x=11$ makes LHS = RHS and is the solution of the given equation.

Linear dependency signifies that "a","b"and "c" are linearly related, which means that if one of them can be found, the other two can be found.

Hence, the given equation can be reduced to a linear equation if the exact relation between "a","b"and "c" is given.

For a better understanding, let us express the dependencies as equations.

Since b is linearly dependent on a, we can write b = ka + p, where k and p are some constants. Note that in this equation, b and a are the only variables, whereas k and p are known.

Similarly, a can be written as a = gc + u, where g and u are constants (known values).

Substituting these values of a and b in the given equation, we get,
3a + 2b + 4c = 5
3(gc + u) + 2(ka + p) + 4c = 5
3gc + 3u + 2p + 2k(gc + u) + 4c = 5
(3g + 2kg + 4)c + (3u + 2p + 2ku - 5) = 0.

Note that in the final equation, the only c is unknown, the others are all constants and hence is a linear equation in one variable. 

Given,
$\frac{x+5}{6}-\frac{x+1}{9}=\frac{x+3}{4}$ 

Multiplying both sides by 36 (LCM of 4, 6 and 9), we get,

$(\frac{x+5}{6}\times 36)-(\frac{x+1}{9}\times 36)=(\frac{x+3}{4}\times 36)$ 

$\Rightarrow 6(x+5)–4(x+1)=9(x+3)$
$\Rightarrow 6x+30–4x-4=9x+27$
$\Rightarrow 2x+26=9x+27$

Transposing 9x from RHS to LHS, and 26 from LHS to RHS, we get,

$2x-9x=27-26$
$\Rightarrow -7x=1$

Dividing both sides by -7, we get,
$x=-\frac{1}{7}$

Number of 1000 denomination notes with Pawan = $2x$
Therefore, amount = $₹2000x$

Number of 500 denomination notes with Pawan  = $3x$
Therefore, amount = $₹1500x$

Number of 100 denomination notes with Pawan = $5x$
Therefore, amount = $₹500x$

Total amount = $₹5,00,000$ (Given)

$2000x+1500x+500x=₹5,00,000$ 

                           $4000x=5,00,000$

                            $x=\frac{5,00,000}{4,000}=125$

Therefore, number of 1000 denomination notes = $2x=2\times 125$
= $250$ notes.

$\frac{3}{x+1}=\frac{6}{5x-1}$

On cross multiplication, we get
$\Rightarrow 3(5x-1)=6(x+1)\Rightarrow 15x-3=6x+6\Rightarrow 9x=9\Rightarrow x=1$

Let the number be $x$.
'Subract  ¼ from the number' is mathematically written as 'x - ¼'.
Multiplying the above obtained result by ½ is mathematically shown as:
'(x - ¼) × ½'.
According to the question,
$(x-\frac{1}{4})\times \frac{1}{2}=\frac{1}{4}$
$\Rightarrow x-\frac{1}{4}=2\times \frac{1}{4}$ 
                    (by multiplying both the sides by 2)
$\Rightarrow x-\frac{1}{4}=\frac{1}{2}$
$\Rightarrow x=\frac{1}{2}+\frac{1}{4}$
                    (by transposing ¼ from LHS to the RHS)
$\Rightarrow x=\frac{3}{4}$

So, the required number is $\frac{3}{4}$.