Question
Solve the following equations by trial and error method: $x + 3 = 12$
✓
Answer
$x + 3 = 12$ Here, $L.H.S. = x + 3$ and $R.H.S. = 12$
Therefore, if $x = 9, L.H.S. = R.H.S$. Hence, $x = 9$ is the solution to this equation.
|
x
|
L.H.S.
|
R.H.S.
|
Is L.H.S. = R.H.S.
|
|
$1$
|
$1 + 3 = 4$
|
$12$
|
No
|
|
$2$
|
$2 + 3 = 5$
|
$12$
|
No
|
|
$3$
|
$3 + 3 = 6$
|
$12$
|
No
|
|
$4$
|
$4 + 3 = 7$
|
$12$
|
No
|
|
$5$
|
$5 + 3 = 8$
|
$12$
|
No
|
|
$6$
|
$6 + 3 = 9$
|
$12$
|
No
|
|
$7$
|
$7 + 3 = 10$
|
$12$
|
No
|
|
$8$
|
$8 + 3 = 11$
|
$12$
|
No
|
|
$9$
|
$9 + 3 = 12$
|
$12$
|
Yes
|
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
By what number should we multiply $\frac{-15}{28}$ so that the product may be $\frac{-5}{7}?$
→Find the area of a rectangular field in hectares whose sides are: $75m$ $5dm$ and $120m$
→→
How much is $x^2-2 x y+3 y^2$ less than $2 x^2-3 y^2+x y$ ?
→The lengths of the sides of triangles are given below. Which of them are right-angled? $a = 10\ cm, b = 24\ cm, c = 26\ cm$
→The adjacent sides of a parallelogram are 10 m and 8 m. If the distance between the longer sides is 4 m, find the distance between the shorter sides.
A horse is tethered to one corner of a rectangular field, $60m$ by $40m$, by a rope $14m$ long. On how much area can the horse graze?
→A father is $30$ years older than his son. In $12$ years. The man will be three times as old as his son. Find their present ages.
→The price of 1 kg of oranges is ₹ 56.50. What is the price of 2.250 kg of oranges? Can we write 56.50 as 56.5 and 2.250 as 2.25 and multiply? Will we get the same product? Why?
Simplify: $2 \frac{2}{17} \times 7 \frac{2}{9} \times 1 \frac{33}{52}$
→