Question

Solve the following equations by hit and trial method:

$3\mathrm{y}-1=14$

Answer 1

Given equation: $3\mathrm{y}-1=14$

The side on the left of this equal sign $(=)$ is called the left hand side (or L.H.S.) and the one on the right is called the right hand side (or R.H.S.)

so, here L.H.S $=$$3\mathrm{y}-1$ and R.H.S $=14$

Now try $\mathrm{y}=1,2,3,4,5...$

When, $\mathrm{y}=1$ L.H.S $=3\times 1-1=2\ne$R.H.S

When, $\mathrm{y}=2$ L.H.S $=3\times 2-1=5\ne$R.H.S

When, $\mathrm{y}=1$ L.H.S $=3\times 3-1=8\ne$R.H.S

When, $\mathrm{y}=2$ L.H.S $=3\times 4-1=11\ne$R.H.S

When, $\mathrm{y}=1$ L.H.S $=3\times 5-1=14=$R.H.S

Hence, $\mathrm{y}=5$ is the solution.