Question

Solve each of the following equations by trial-and-error method:
(i) x + 3 = 12
(ii) x − 7 = 10
(iii) 4x = 28
(iv) $\frac{x}{2}+7=11$
(v) 2x + 4 = 3x
(vi) $\frac{x}{4}=12$
(vii) $\frac{15}{x}=3$
(viii) $\frac{x}{18}=20$

Answer 1

(i) x + 3 = 12
Here, LHS = x + 3 and RHS = 12.

x	LHS	RHS	Is LHS = RHS?
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

Therefore, if x = 9, LHS = RHS.
Hence, x = 9 is the solution to this equation.

(ii) x − 7 = 10
Here, LHS = x −7 and RHS =10.

x	LHS	RHS	Is LHS = RHS?
9	9−7 = 2	10	No
10	10−7 = 3	10	No
11	11−7=4	10	No
12	12−7=5	10	No
13	13−7=6	10	No
14	14−7=7	10	No
15	15−7=8	10	No
16	16−7=9	10	No
17	17−7=10	10	Yes

Therefore, if x = 17, LHS = RHS.
Hence, x = 17 is the solution to this equation.

(iii) 4x = 28
Here, LHS = 4x and RHS = 28.

x	LHS	RHS	Is LHS = RHS?
1	4$\times$1=4	28	No
2	4$\times$2=8	28	No
3	4$\times$3=12	28	No
4	4$\times$4=16	28	No
5	4$\times$5=20	28	No
6	4$\times$6=24	28	No
7	4$\times$7=28	28	Yes

Therefore, if x = 7, LHS = RHS.
Hence, x = 7 is the solution to this equation.

(iv) $\frac{x}{2}+7=11$
Here, LHS = $\frac{x}{2}+7$ and RHS = 11.
Since RHS is a natural number, $\frac{x}{2}$ must also be a natural number, so we must substitute values of x that are multiples of 2.

x	LHS	RHS	Is LHS = RHS?
2	$\frac{2}{2}$+7=8	11	No
4	$\frac{4}{2}$+7=9	11	No
6	$\frac{6}{2}$+7=10	11	No
8	$\frac{8}{2}$+7=11	11	Yes

Therefore, if x = 8, LHS = RHS.
Hence, x = 8 is the solution to this equation.

(v) 2x + 4 = 3x
Here, LHS = 2x + 4 and RHS = 3x.

x	LHS	RHS	Is LHS = RHS?
1	2(1)+4=6	3(1)=3	No
2	2(2)+4=8	3(2)=6	No
3	2(3)+4=10	3(3)=9	No
4	2(4)+4=12	3(4)=12	Yes

Therefore, if x = 4, LHS = RHS.
Hence, x = 4 is the solution to this equation.

(vi) $\frac{x}{4}$ = 12
Here, LHS =$\frac{x}{4}$ and RHS = 12.
Since RHS is a natural number, $\frac{x}{4}$ must also be a natural number, so we must substitute values of x that are multiples of 4.

x	LHS	RHS	Is LHS = RHS?
16	$\frac{16}{4}$=4	12	No
20	$\frac{20}{4}$=5	12	No
24	$\frac{24}{4}$=6	12	No
28	$\frac{28}{4}$=7	12	No
32	$\frac{32}{4}$=8	12	No
36	$\frac{36}{4}$=9	12	No
40	$\frac{40}{4}$=10	12	No
44	$\frac{44}{4}$=11	12	No
48	$\frac{48}{4}$=12	12	Yes

Therefore, if x = 48, LHS = RHS.
Hence, x = 48 is the solution to this equation.

(vii) $\frac{15}{x}$ = 3

Here, LHS =$\frac{15}{x}$ and RHS = 3.
Since RHS is a natural number, $\frac{15}{x}$ must also be a natural number, so we must substitute values of x that are factors of 15.

x	LHS	RHS	Is LHS = RHS?
1	$\frac{15}{1}$=15	3	No
3	$\frac{15}{3}$=5	3	No
5	$\frac{15}{5}$=3	3	Yes

Therefore, if x = 5, LHS = RHS.
Hence, x = 5 is the solution to this equation.

(viii) $\frac{x}{18}$= 20
Here, LHS =$\frac{x}{18}$ and RHS = 20.
Since RHS is a natural number, $\frac{x}{18}$ must also be a natural number, so we must substitute values of x that are multiples of 18.

x	LHS	RHS	Is LHS = RHS?
324	$\frac{324}{18}$=18	20	No
342	$\frac{342}{18}$=19	20	No
360	$\frac{360}{18}$=20	20	Yes

Therefore, if x = 360, LHS = RHS.
Hence, x = 360 is the solution to this equation.