MATHS M.C.Q. [1 Marks Each]

 Let the monthly salary is $Rs.x$ Then pocket expenses is $\frac{1}{5}\text{x}=\frac{\text{x}}{5}$ And other expenses is $\frac{4}{5}$ of reminder $\frac{4\text{x}}{5}=\frac{16}{25}$ Then total expenses $=\frac{\text{x}}{5}+\frac{16\text{x}}{25}=\frac{21\text{x}}{25}$
Then savings $=\text{x} +\frac{21\text{x}}{25}=\frac{4\text{x}}{25}$ But saving given $\text{Rs }1200 \frac{ 4\text{x}}{25}=1200 \Rightarrow\text{x}=7500\text{ Rs}$
Then monthly salary $Rs.7500.$

$ 2n + 5 = 3(3n - 10)$
$⇒ 2n + 5 = 9n - 30$
$⇒ 9n - 2n = 5 + 30$
$⇒ 7n = 35$
$⇒ n = 5$

$ 3s = 0$
$⇒ s = \frac{0}{3}$
$⇒ s = 0 ($if we divide $0$ by any number except $0,$ we will always get $0)$

Given equation is $43m = 0.086$
On dividing the given equation by $43,$ we get
$\text{m}=\frac{0.086}{43}$
If we remove the decimal, we get $1000$ in denominator
$\text{m}=\frac{86}{43}\times\frac{1}{1000}=\frac{1}{1000}=0.002$

$x - 6 = 1$
$\Rightarrow x = 1 + 6 = 7.$

 Let the number is $x$
One-third of $\text{ x} = \frac{\text{x}}{3}$
Given, One-third of a number added to itself gives $10$
$\Rightarrow\frac{\text{x}}{3} + \text{x} = {10}$

 Average score of Maya, Madhura and Mohsina $= 19$
So $($Score of Maya $+$ Score of Madhura $+$ Score of Mohsina$) 3 = 19$
$\Rightarrow $ Score of Maya $+$ Score of Madhura $+$ Score of Mohsina $= 19 \times 3$
$\Rightarrow $ Score of Maya $+$ Score of Madhura $+$ Score of Mohsina $= 57$
$\Rightarrow 16 + 20 +$ Score of Madhura $= 57$
$\Rightarrow 36 +$ Score of Madhura $= 57$
$\Rightarrow $ Score of Madhura $= 57 - 36$
$\Rightarrow $ Score of Madhura $= 21$

 The given statement means $x$ is $7$ more than $3.$
So, the equation is $x - 7 = 3$
We can also write it as $x - 3 = 7.$

Let the number be $x.$
As, two-third of a number is greater than one-third of the number by $5.$
$\Rightarrow\frac{2}{3}\text{x}-\frac13\text{x}=5$
$\Rightarrow\frac{2\text{x}-\text{x}}{3}=5$
$\Rightarrow\frac{\text{x}}{3}=5$
$\Rightarrow\text{x}=5\times3$
$\therefore\text{x}=15$
So, the number is $15.$
Hence, the correct alternative is option $(c).$

 Let the two consecutive odd numbers be $x$ and $x + 2.$
As, the sum of the two consecutive odd numbers is $36.$
$\Rightarrow x + (x + 2) = 36$
$\Rightarrow 2x + 2 = 36$
$\Rightarrow 2x = 36 - 2$
$\Rightarrow 2x = 34$
$\Rightarrow\text{x}=\frac{34}2$
$\Rightarrow x = 17$
$\therefore x + 2 = 17 + 2 = 19$
So, the larger number is $19.$
Hence, the correct alternative is option $(c).$

Given, $\frac{\text{x}}{3}=4$ then the value of $2x + 5$ is
$\Rightarrow x = 4 \times 3$
$\Rightarrow x = 12$
Now, $2x + 5 = 2 \times 12 + 5 = 24 + 5 = 29$

$4x + 5 = 9$
$\Rightarrow 4x = 9 - 6 = 4$
$\Rightarrow\text{x} = \frac{4}{4} = 1$

 $\frac{2\text{x}-1}{3}=\frac{\text{x}-2}{3}+1$
$\Rightarrow\frac{2\text{x}-1=\text{x}-2 +3}{3}$
$\Rightarrow2\text{x}-\text{x}=-2+3+1$
$\Rightarrow\text{x}=2$
$\therefore\text{x}=2$

Given, $x$ exceed 4 by $9$
$\Rightarrow x = 9 + 4$
$\Rightarrow x - 4 = 9$

$5x = 10$
$\Rightarrow\text{x} = \frac{10}{5} = 2 $

Given, Add $9$ to $5$ times $n$ to get $3$
Now, $5$ times of $n = 5n$
Add $9,$ we get
$5n + 9$
This is equal to $3$
So, $5n + 9 = 3$

Let first whole number $= x$
Then second number $= x + 1$
And sum $= 53$
$x + x + 1 = 53$
$⇒ 2x = 53 - 1$
$⇒ 2x = 52$
$⇒ x = 26$
Smaller number $= 26$

$ 4p - 3 = 9$
$\Rightarrow 4p = 9 + 3 = 12$
$\Rightarrow\text{p} = \frac{12}{4} = 3.$

$\Rightarrow 10y + 20 = 50$
$\Rightarrow 10y = 50 - 20$
$\Rightarrow 10y = 30$
$ = \frac{30}{10}$
$\Rightarrow y = 3$

Given, $43x = 0.086$
${43}\text{x} = \frac{86}{1000}$
$\Rightarrow\text{x} = \frac {86}{(1000 \times43)}$
$\Rightarrow\text{ x} = \frac{2}{1000}$
$\text{x} = \frac{1}{500}$

Let the number be $x.$
As, the sum of a number and its two-fifth is $70.$
$\Rightarrow\text{x}+\frac25\text{x}=70$
$\Rightarrow\frac{\text{x}}1+\frac{\text{2x}}5=70$
$\Rightarrow\frac{\text{5x}}{5}+\frac{\text{2x}}{5}=70$
$\Rightarrow\frac{5\text{x}+2\text{x}}{5}=70$
$\Rightarrow\frac{7\text{x}}{5}=70$
$\Rightarrow7\text{x}=70\times5 ($By transposing $5$ to $R.H.S.)$
$\Rightarrow7\text{x}=350$
$\Rightarrow\text{x}=\frac{350}{7} ($By transposing $7$ to $R.H.S.)$
$\therefore\text{x}=50$
So, the number is $50.$
Hence, the correct alternative is option $(b).$

Let first odd number $= 2x + 1$
Second number $= 2x + 3$
$2x + 1 + 2x + 3 = 36$
$⇒ 4x + 4 = 36$
$⇒ 4x = 36 - 4 = 32$
$⇒ x = 8$
Smaller number $= 2x + 1 = 2 × 8 + 1 = 16 + 1 = 17$

$\frac{2}{3}\text{x} + 6 = 12$
$\Rightarrow\frac{2}{3}\text{x}=12-6$
$\Rightarrow\frac{2}{3}\text{x}=6$
$⇒ 2\text{x} = 6\text { x} 3$
$⇒\text{x}=\frac{6×3}{2}$
$⇒\text{x}=9$

Let $2$ number be $x$ and $y.$
So larger number be $x.$
Given difference of $2$ numbers is $21$
i.e larger number minus smaller number is $21 $
$​​​​​​​⇒ x - y = 21$
So the smaller number $y$ is given as $x - 21$

$3s + 12 = 0$
$\Rightarrow 3s = 0 - 12$
$\Rightarrow 3s = -12$
$\Rightarrow\text{s} = \frac{-12}{3}$
$\Rightarrow s = -4$

$\therefore\frac{2}{3}\text{x}=\text{x}-10$
$\Rightarrow\text{x}-\frac{2}{3}\text{x}=10$
$\Rightarrow\frac{1}{3}\text{x}=10$
$\Rightarrow\text{x}=30$

We have, $x = 0$
On multiplying both the sides by $3,$ we get
$3 \times x = 3 \times 0$
$\Rightarrow 3x = 0$
On adding $(-1)$ both the sides, we get
$3x + (-1) = 0 + (-1)$
$\Rightarrow 3x - 1 = -1$

two times $y = 2y$
Now the sum of two times $y$ and $10$ is $42$ is written as $2y + 10 = 42$

Subtraction $3$ from $2x = 2x - 3$

$2\text{z}+\frac{8}{3}=\frac{1}{4}\text{z}+5$
$\Rightarrow2\text{z}-\frac{1}{4}\text{z}=5-\frac{8}{3}$
$\Rightarrow\frac{8\text{z}-\text{z}}{4}=\frac{15-8}{3}$
$\Rightarrow\frac{7}{4}\text{z}=\frac{7}{3}$
$\Rightarrow\text{z}=\frac{7}{3}\times\frac{4}{7}$
$=\frac{4}{3}$

Given, $\frac{\text{x}}{3} = 4,$ then the value of $2x + 5$ is
$\Rightarrow x = 4 \times 3$
$\Rightarrow x = 12$
Now, $2x + 5 = 2 \times 12 + 5 = 24 + 5 = 29$

Let first angle $= x$
Then second $= 90^\circ - x$
$x - (90^\circ - x) = 10$
$\Rightarrow x - 90^\circ + x = 10^\circ $
$\Rightarrow 2x = 10^\circ + 90^\circ = 100^\circ $
$x = 50^\circ $
Second angle $= 90^\circ - 50^\circ = 40^\circ $
Larger angle $= 50^\circ $

Let the multiples of $2$ are $x$ and $x + 2$
Given, sum $= 18$
$\Rightarrow x + x + 2 = 18$
$\Rightarrow 2x + 2 = 18$
$\Rightarrow 2x = 18 - 2$
$\Rightarrow 2x = 16$
$\Rightarrow\text{ x} = \frac{16}{2}$
$\Rightarrow x = 8$
So, the numbers are $8, 8 + 2 = 8, 10$

$10y - 20 = 50$
$\Rightarrow 10y = 50 + 20$
$\Rightarrow 10y = 70$
$\Rightarrow\text{y} = \frac{70}{10}$
$\Rightarrow y = 7$

Given, $x = 2,$
Now $\frac{(1 - 3\text{x})}{3} = \frac{({1-3\times2)}}{3}$
$= \frac{(1−6)}{3}$
$= \frac{(−5)}{3}$

Let the larger angle be $x.$
Then, the smaller angle $= (x - 40^\circ )$
As, the sum of the two supplementary angles is always $180^\circ .$
$\Rightarrow x + (x - 40^\circ ) = 180^\circ $
$\Rightarrow 2x - 40^\circ = 180^\circ $
$\Rightarrow 2x = 180^\circ + 40^\circ $
$\Rightarrow 2x = 220^\circ $
$\Rightarrow\text{x}=\frac{220^\circ}{2} ($By transposing $2$ to $R.H.S.)$
$\therefore\text{x}=1106^\circ$
So, the measure of the larger angle is $110^\circ .$
Hence, the correct alternative is option $(c).$

Given, $5x - 8 = x + 4$
$\Rightarrow 5x - 8 - x = 4$
$\Rightarrow 4x - 8 = 4$
$\Rightarrow 4x = 4 + 8$
$\Rightarrow 4x = 12$
$\Rightarrow\text{x} = \frac{12}{4}$
$\Rightarrow x = 3$

$1400 \times\text{x} =1050$
$\Rightarrow{\text{x}}=\frac{1050}{1400}=\frac{3}{4}$

$ 7n + 5 = 12$
$\Rightarrow 7n = 12 - 5$
$\Rightarrow 7n = 7$
$\Rightarrow \text{n}= \frac{7}{7}$
$= 1$

Given equation is $ax = b$
On dividing the equation by $a,$ we get
$\text{x}=\frac{\text{b}}{\text{a}}$
Now, if $a$ and $b$ are positive integers, then the solution of the equation is also positive number as division of two positive integers is also a positive number.

Any even number is of the form $x = 2n$ where $n$ is an integer So then consecutive even number will be $2n + 2$ i.e $x + 2$

Let the number be $x.$
As, $23$ of the number is less than the original number by $20.$
$\Rightarrow\text{x}-\frac23\text{x}=20$
$\Rightarrow\frac{\text{x}}{1}-\frac{\text{2x}}{3}=20$
$\Rightarrow\frac{\text{3x}}{\text{3}}-\frac{\text{2x}}{3}=20$
$\Rightarrow\frac{\text{3x}-\text{2x}}{3}=20$
$\Rightarrow\frac{\text{x}}{3}=20$
$\Rightarrow\text{x}=20\times3 ($By transposing $3$ to $R.H.S.)$
 $\therefore\text{x}=60$
So, the number is $60.$
Hence, the correct alternative is option $(d).$

$2\text{x}+\frac{1}{3\text{x}}=15$ Multiply both side by $3x$ Then So,
$\frac{5\text{x}}{6\text{x}^2+20\text{x}+1}=\frac{5\text{x}}{\text{6x}^2-15\text{x}+1+\text{35x}}=\frac{5\text{x}}{0+\text{35x}}\Rightarrow\frac{5\text{x}}{35\text{x}}=\frac{1}{7}$

Let number $= x$ then
$3x + 6 = 24$
$\Rightarrow 3x = 24 - 6 = 18$
$\Rightarrow x = 6$
Number $= 6$

 Given equation is $7x + 4 = 25$
$⇒ 7x = 25 - 4 [$transposing $4$ to $RHS]$
$⇒ 7x = 21$
On dividing the above equation by $7,$ we get
$x = 3$
Hence, the solution of the given equation is $3.$

 Let the age of Sohan $= x$
Now, age of Mohan $= x + 3$
Given, sum of their ages $=$
$\Rightarrow x + x + 3 = 43$
$\Rightarrow 2x + 3 = 43$
$\Rightarrow 2x = 43 - 3$
$\Rightarrow 2x = 40$
$\Rightarrow\text{x} = \frac{40}{2}$
$\Rightarrow x = 20$
So, age of Sohan is $20$ years

Given, $a$ and $b$ are positive integers
Again, $ax = b$
$ \Rightarrow \text{x} = \frac{\text{b}}{\text{a}}$
Since $a$ and $b$ are positive integers, So $ \frac{\text{b}}{\text{a}}$ is also a positive number.

Given, $\frac{\text{x}}{2}=3$
On muliplying both sides by 2, we get $\frac{\text{x}}{2}\times2=3\times2$
$\Rightarrow\text{x}=3\times2=6$
Put $x = 6$ in the equation $3x + 2,$ we get
$3(6) + 2 = 18 + 2 = 20$

$m - 1 = 2$
$\Rightarrow m = 2 + 1 = 3.$

Let the two complementary angles be $x^\circ $ and $(90 - x)^\circ .$
According to the equation, we have:
$x - (90 - x) = 14$
$\Rightarrow 2x = 104$
$\Rightarrow x = 52$
$(90^\circ - x)^\circ = 90^\circ - 52^\circ = 38^\circ $
The larger angle is $52^\circ .$

As, $\frac{\text{x}+2}{\text{x}-2}=\frac{2}{3}$
$\Rightarrow3(\text{x}+2)=2(\text{x}-2)$ (By cross multiplication)
$\Rightarrow\text{3x}+6=2\text{x}-4$
$\Rightarrow\text{3x}-\text{2x}=-6+4 ($By transposing $2x$ to $L.H.S.$ and $6$ to $R.H.S.)$
$\therefore\text{x}=-10$
Hence, the correct alternative is option $(a).$

$10y - 20 = 30$
$\Rightarrow 10y = 30 + 20 = 50$
$\Rightarrow\text{y} = \frac{50}{10} = 5$

$\frac{\text{x}-1}{\text{x}+1}=\frac{7}{9}$
$\Rightarrow9\text{x}-9=7\text{x}+7$
$\Rightarrow9\text{x}-7\text{x}=7+9$
$\Rightarrow2\text{x}=16$
$\Rightarrow\text{x}=\frac{16}{2}=8$

Given equation is $ax + b = 0$
$\Rightarrow\text{ax}=-\text{b}[$ transposing b to $RHS]$
$\Rightarrow\text{x}=-\frac{\text{b}}{\text{a}} [$on dividing both sides by $a]$

$ x + 3 = 0$
$\Rightarrow x = -3.$

 As, $\frac{\text{x}}{6}+\frac{\text{x}}{4}=\frac{\text{x}}{2}+\frac34$
$\Rightarrow\frac{\text{x}}{6}+\frac{\text{x}}{4}-\frac{\text{x}}{2}=\frac{3}{4} ($By transposing $\frac{\text{x}}{2}$ to $L.H.S.)$
$\Rightarrow\frac{2\text{x}}{12}+\frac{3\text{x}}{12}-\frac{6\text{x}}{12}=\frac{3}{4}$
$\Rightarrow\frac{2\text{x}+3\text{x}-6\text{x}}{12}=\frac34$
$\Rightarrow\frac{-\text{x}}{12}=\frac34$
$\Rightarrow-\text{x}\times4=3\times12$ (By cross multiplication)
$\Rightarrow-4\text{x}=36$
$\Rightarrow\text{x}=\frac{36}{-4}$
$\therefore\text{x}=-9$
Hence, the correct alternative is option $(c).$

Consider the given intersects $2$ and $3$ on $x-$axis $x−$axis and $y-$axis $y−$axis respectively.
$\text{a}=\pm2,\text{b}=\pm3$
We know that the equation of the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
So, $\frac{\text{x}}{\pm2}+\frac{\text{y}}{\pm3}-1$
$\pm3{\text{x}}\pm2{\text{y}}=6$
Hence, this is the answer.

$p + 4 = 4$
$\Rightarrow P = 4 - 4 = 0.$

Given, $\frac{\text{x}}{5}-2=6$
$\Rightarrow \frac{\text{x}}{5}=6+{2}$
$\Rightarrow \frac{\text{x}}{5}={8}$
$\Rightarrow x = 8 \times 5$
$\Rightarrow x = 40$

$\Rightarrow $ Given, $7x - 4 = -25$
$\Rightarrow 7x = -25 + 4$
$\Rightarrow 7x = -21$
$\Rightarrow\text{x} = \frac{-21}{7}$
$\Rightarrow x = -3$

Let the number is $x$
Given, if a number is increased by $25,$ it becomes $40$
$\Rightarrow x + 25 = 40$
$\Rightarrow x = 40 - 25$
$\Rightarrow x = 15$

$3p + 5 = 8$
$\Rightarrow 3p = 8 - 5 = 3$

Let the number is $x$
Given, $10$ less than a number is $55$
$\Rightarrow x - 10 = 55$
$\Rightarrow x = 55 + 10$
$\Rightarrow x = 65$

$3m + 7 = 16$
$\Rightarrow 3m = 16 - 7 = 9$
$\Rightarrow\text{m} =\frac{ 9}{3} = 3$

Given, $\frac{\text{x}}{3} = 4$ then the value of $2x + 5$ is
$\Rightarrow x = 4 \times 3$
$\Rightarrow x = 12$
Now, $2x + 5 = 2 \times 12 + 5 = 24 + 5 = 29$

$4p - 2 = 10$
$\Rightarrow 4p = 10 + 2 = 12$
$\Rightarrow\text{p} = \frac{12}{4} = 3$

Let the number be $x.$
As, twice the number when increased by $7$ gives $25.$
$\Rightarrow 2x + 7 = 25$
$\Rightarrow 2x = 25 - 7 ($By transposing $7$ to $R.H.S.)$
$\Rightarrow 2x = 18$
$\Rightarrow\text{x}=\frac{18}{2} ($By transposing $2$ to $R.H.S.)$
$\therefore\text{x}=9$
So, the number is $9.$
Hence, the correct alternative is option $(b).$

$y + 2 = -2$
$\Rightarrow y = -2 - 2 = -4$

$-4 (2 - x) = 9$
$\Rightarrow -4 \times 2 + 4 \times x = 9$
$\Rightarrow -8 + 4x = 9$
$\Rightarrow 4x = 9 + 8$
$\Rightarrow 4x = 17$
$\Rightarrow\text{x}=\frac{17}{4}$

Let first even number $= 2x$
Then second number $= 2x + 2$
And sum $= 86$
$2x + 2x + 2 - 86$
$\Rightarrow 4x = 86 - 2 = 84$
$\Rightarrow x = 21$
Larger even number $= 2x + 2 = 2 \times 21 + 2 = 42 + 2 = 44$

Let the number $= x$
According to the condition,
$5x = 80 + x$
$⇒ 5x - x = 80$
$⇒ 4x = 80$
$⇒ x = 20$
Number $= 20$

$5\text{x} = 10$
$\Rightarrow\text{x} = \frac{10}{5}$
$= 2$

$8(2x - 5) - 6(3x - 7) = 1$
$⇒ 16x - 40 - 18x + 42 = 1$
$⇒ -2x + 2 = 1$
$⇒ -2x = 1 - 2 = -1$
$\text{x}=\frac{1}{2}$

$x - 6 = 1$
$⇒ x = 1 + 6$
$= 7$

As, $\frac{\text{x}-2}{3}=\frac{2\text{x}-1}{3}-1$
$\Rightarrow\frac{\text{x}-2}{3}-\frac{2\text{x}-1}{3}=-1 ($By transposing $\frac{2\text{x}-1}{3}$ to $L.H.S.)$
$\Rightarrow\frac{(\text{x}-2)-(\text{2x}-1)}{3}=-1$
$\Rightarrow\frac{\text{x}-2-2\text{x}+1}{3}=-1$
$\Rightarrow\frac{-\text{x}-1}{3}=-1$
$\Rightarrow-\text{x}-1=-1\times3( $By transposing $3$ to $R.H.S.)$
$\Rightarrow-\text{x}-1=-3$
$\Rightarrow-\text{x}=-3-1 ($By transposing $-1$ to $R.H.S.)$
$\Rightarrow-\text{x}=-2$
$\therefore\text{x}=2$
Hence, the correct alternative is option $(a).$

In a equation, before equal is called Left-hand side $(LHS)$ and after equal is called Right-hand side $(RHS)$
So in Equation $3x + 4 = 25,$
$RHS$ is $25.$

Let the number be $x.$
According to the equation, we have:
$4x = x + 54$
$\Rightarrow 3x = 54$
$\Rightarrow x = 18$

$3s + 12 = 0$
$\Rightarrow 3s = 0 - 12$
$\Rightarrow 3s = -12$
$\Rightarrow\text{s} = \frac{-12}{3}$
$\Rightarrow s = -4$

As, $2\text{x}-\frac32=\text{5x}+\frac34$
$\Rightarrow2\text{x}-5\text{x}=\frac{3}{4}+\frac{3}{4} ($By transpoing $-\frac32$ to $R.H.S.$ and $5x$ to $L.H.S.)$
$\Rightarrow-3\text{x}=\frac{6}{4}+\frac34$
$\Rightarrow-3\text{x}=\frac{6+3}{4}$
$\Rightarrow\text{x}=\frac{9}{4\times(-3)} ($By transposing $-3$ to $R.H.S.)$
$\Rightarrow\text{x}=\frac{3}{4\times(-1)}$
$\Rightarrow\text{x}=\frac{3}{-4}$
$\therefore\text{x}=-\frac34$
Hence, the correct alternative is option $(b).$

As, $2(2n + 5) = 3(3n - 10)$
$⇒ 4n + 10 = 9n - 30$
$⇒ 4n - 9n = -10 - 30 ($By transposing $10$ to $R.H.S.$ and $9n$ to $L.H.S.)$
$⇒ -5n = -40$
$⇒ n = -40 - 5 ($By transposing $-5$ to $R.H.S.)$
$\therefore\text{n}=8$
Hence, the correct alternative is option $ (d).$

 $\frac{\text{x}}{2}-1=\frac{\text{x}}{3}+4$
$\Rightarrow\frac{\text{x}}{2}-\frac{\text{x}}{3}=4+1$
$\Rightarrow\frac{3\text{x}-2\text{x}}{6}=5$
$\Rightarrow\frac{\text{x}}{6}=5$
$\Rightarrow\text{x}=5\times6=30$
$\therefore\text{x}=30$

$2p - 1 = 3$
$\Rightarrow 2p = 3 + 1 = 4$
$\Rightarrow\text{p} = \frac{4}{2} = 2$

$\frac{\text{x}}{2}-\frac{\text{x}}{3}=5$
$\Rightarrow\frac{3\text{x}-2\text{x}}{6}=5$
$\Rightarrow\text{x}=30$

Given
$\Rightarrow\frac{\text{x}}{3} = \frac{5}{4}$
$\Rightarrow $ Apply cross-multiplication, we get
$\Rightarrow x \times 4 = 5 \times 3$
$\Rightarrow 4x = 15$
$\text{x} = \frac{15}{4}$

 Let the number is $x$
Given, the sum of twice a number and $4$ is $18$
$\Rightarrow 2x + 4 = 18$
$\Rightarrow 2x = 18 - 4$
$\Rightarrow 2x = 14$
$\Rightarrow x = \frac{14}{2}$
$\Rightarrow x = 7$

We have, $x = 7$
On multiplying both the sides by $7,$ we get
$7 \times x = 7 \times 7 $
$\Rightarrow 7x = 49$
On adding $(-1)$ both the sides, we get
​​​​​​​$7x + (-1) = 49 + (-1)$
$\Rightarrow 7x - 1 = 49 - 1$
$\Rightarrow 7x - 1 = 48$