Question
Solve the following equation and verify the answer: $6(1 - 4x) + 7(2 + 5x) = 53$

Answer

$6(1 - 4 x) + 7(2 + 5x) - 53 $
$\Rightarrow 6 - 24x + 14 + 35x = 53$ (Removing brackets)
$\Rightarrow -24x + 35x + 14 + 6 = 53 $
$\Rightarrow 11x + 20 = 53 $
$\Rightarrow 11x = 53 - 20 $
$\Rightarrow 11x = 33$ (Transposing $20$ to $R.H.S.)$
$\Rightarrow\frac{\text{11x}}{11}=\frac{33}{11}$(Dividing both sides by $11)$
$\Rightarrow x = 3$
So, $x = 3$ is a solution of the given equation.
Check: Substituting $x = 3$ in the given equation,
we get, $L.H.S. = 6 (1 - 4 \times 3) + 7(2 + 5 \times 3)$
$= 6(1 - 12) + 7(2 + 15)$
$= 6 \times (-11) + 7 \times 17$
$= -66 + 119 = 53$
$= R.H.S.$
$\therefore$ When $x = 3,$
 we have $L.H.S. = R.H.S.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Following is the list of temperatures of five places in India on a particular day of the year.
Place Temperature  
Siachin $10^\circ C$ below $0^\circ C$ _________
Shimla $2^\circ C$ below $0^\circ C$ _________
Ahmedabad $30^\circ C$ above $0^\circ C$ _________
Delhi $20^\circ C$ above $0^\circ C$ _________
Srinagar $5^\circ C$ below $0^\circ C$ _________
Which is the coolest place$?$
Which direction will you face if you start facing east and make $\frac{1}{2}$ of a revolution clockwise?
Fill in the missing fraction: $\square+\frac5{27}\;=\;\frac{12}{27}$
Draw number line and locate the points on them: $\frac18,\;\frac28,\frac38,\;\frac78$.
Construct a rectangle one of whose sides is 4 cm and the diagonal is of length 8 cm.
The sum of two positive integers is always positive but a (positive integer) – (positive integer) can be positive or negative. What about
(a) (Positive) – (Negative)
(b) (Positive) + (Negative)
(c) (Negative) + (Negative)
(d) (Negative) – (Negative)
Use a pair of compasses and construct the following angles: $45^\circ $
Reduce the fraction to simplest from: $\frac{12}{52}$
The weights of new born babies (in kg) in a hospital on a particular day are as follows:
2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1, 2.5, 2.8, 2.7, 2.9,
(i) Rearrange the weights in descending order.
(ii) Determine the highest weight.
(iii) Determine the lowest weight.
(iv) Determine the range.
(v) How many babies were born on that day?
(vi) How many babies weigh below 2.5 kg?
(vii) How many babies weigh more than 2.8 kg?
(viii) How many babies weigh 2.8 kg?