76 Basic Engineering Mathematics
i.e.^38 y=−^114Dividing both sides by 38 gives38 y
38=− 114
38
Cancelling gives y=− 3which is the solution of the equation
2 y
5+3
4+ 5 =1
20−3 y
2Problem 12. Solve the equation√
x= 2Whenever square root signs are involved in an equation,
both sides of the equation must be squared.Squaring both sides gives
(√
x) 2
=( 2 )^2i.e. x= 4which is the solution of the equation√
x=2.Problem 13. Solve the equation 2√
d= 8Whenever square roots are involved in an equation, the
square root term needs to be isolated on its own before
squaring both sides.
Cross-multiplying gives √d=^8
2
Cancelling gives√
d= 4
Squaring both sides gives(√
d) 2
=( 4 )^2
i.e. d= 16which is the solution of the equation 2√
d=8.Problem 14. Solve the equation(√
b+ 3
√
b)
= 2Cross-multiplying gives√
b+ 3 = 2√
b
Rearranging gives 3 = 2√
b−√
b
i.e. 3 =√
b
Squaring both sides gives 9 =bwhich is the solution of the equation(√
b+ 3
√
b)
=2.Problem 15. Solve the equationx^2 = 25Whenever a square term is involved, the square root of
both sides of the equation must be taken.Taking the square root of both sides gives√
x^2 =√
25i.e. x=± 5which is the solution of the equationx^2 =25.Problem 16. Solve the equation15
4 t^2=2
3We need to rearrange the equation to get thet^2 term on
its own.Cross-multiplying gives 15 ( 3 )= 2 ( 4 t^2 )i.e. 45 = 8 t^2Dividing both sides by 8 gives 45
8=
8 t^2
8
By cancelling^5.^625 =t^2
or t^2 = 5. 625Taking the square root of both sides gives
√
t^2 =√
5. 625i.e. t=± 2. 372correct to 4 significant figures, which is the solution of
the equation15
4 t^2=2
3Now try the following Practice ExercisePracticeExercise 43 Solving simple
equations (answers on page 344)Solve the following equations.1.1
5d+ 3 = 4- 2+
3
4
y= 1 +2
3
y+5
63.1
4( 2 x− 1 )+ 3 =1
24.1
5( 2 f− 3 )+1
6(f− 4 )+2
15= 05.1
3( 3 m− 6 )−1
4( 5 m+ 4 )+1
5( 2 m− 9 )=− 36.x
3−x
5= 2- 1−
y
3= 3 +y
3−y
6