# 7th Class Mathematics Linear Equations Linear Equation

## Linear Equation

Category : 7th Class

### Introduction

We know that a mathematical statement of equality which involves one or more than one variables is called an equation.

An equation in which variables are of one degree is called linear equation. If there Is only one variable, then the equation is said to be linear equation in one variable.

### Linear Equation

The general form of linear equation in one variable is $ax+b=0,$ where a and b are constant.

The general form of linear equation in two variable is $ax+by+c=0,$ where a, b, c are constants, for example $4x+5=5$ is a linear equation in one variable.

Solutions of the Linear Equations

The real number which satisfies the given equation is called the solution of the equation, for example 6 is the root of the equation $3x+4=5x-8,$ because when we put x = 6 then L.H.S = 22, R.H.S = 22. That is why, to find the solution of an equation means that you have to find the value of variable for which L.H.S = R.H.S The following are the methods to solve a linear equation:

Trial and Error Method

In this method we just guess the roots of the equation.

The value of variable for which L.H.S = R.H.S is the solution of the equation. v+3

e.g. $\frac{y+3}{3}+8=11,$ we guess different values of y, suppose 3, 6, 9 etc.

If we put y = 3 then L.H.S. = 10 and R.H.S = 11.

Therefore, L.H.S$\ne$ R.H.S. hence y = 3 is not a solution. Put y = 6 then L.H.S. = 11 and R.H.S. = 11. Therefore, L.H.S. = R.H.S. hence y = 6 is the solution of given equation

Systematic Method

•    Add or subtract same number from both sides of the equation
•    Multiply or divide both sides of the equation by the same non-zero number.

Solve $19x+2=40$

Solution:

$19x+2=40$ $\Rightarrow 19x+2-2=40-2$(subtracting 2 from both sides)

$\Rightarrow 19x+0=38\Rightarrow 19x=38$

$\Rightarrow \frac{19x}{19}=\frac{38}{19}(dividing\text{ }both\text{ }sides\text{ }by\text{ }19)$

$\Rightarrow x=2.$Therefore, $x=2$is the solution of given equation.

Transposition Method

In this method the following steps are involved:

Step 1:   Identify the unknown quantity

Step 2:   Transfer all the term containing variable on the left hand side and constant term on the right hand side.

Step 3:   Simplify left hand side and right hand side in such a way that each side containing only one term.

Step 4:   Solve the equation obtained above by systematic method.

Solve $3(y+3)-2\text{ (}y-1\text{)}=5(y-5)$

Solution:

$3(y+3)-2(y-1)=5(y-5)$

$\Rightarrow 3y+9-2y+2=5y-25$(Expanding the bracket)

$\Rightarrow y+11=5y-25$

$\Rightarrow y-5y=-25-11$      (Transposing 5y on LHS and 11 on RHS)

$\Rightarrow 4y=-36\Rightarrow \frac{-4y}{-4}=\frac{-36}{-4}=9$

Therefore, y = 9 is the solution of given equation.

The solution of the equation $2(3x-7)+4(3x+2)=6(5x+9)+3$ is:

(a) A natural number

(b) A rational number

(c) An integer

(d) A fraction

(e) None of these

Explanation

We have: $2(3x-7)+4(3x+2)=6(5x+9)+3$

$\Rightarrow 6x-14+12x+8=30x+54+3$[Group like term each side]

$\Rightarrow 6x+12x-30x=54+3+14-8$

$\Rightarrow 18x-30x=63\Rightarrow -12x=63$

$\therefore x=\frac{-63}{12}=\frac{-21}{4}$ is the solution of the equation which is a rational number.

The Solution of the equation $\frac{x}{2}+\frac{1}{2}=\frac{x}{3}-\frac{1}{3}$ represents:

(a) An integer which is less than $-\text{ }4$

(b) An integer which is less than 0

(c) An integer which is between 0 and $-10$

(d) An integer which is equal to $-5$

(e) All of the above

Explanation

$\frac{x}{2}+\frac{1}{2}=\frac{x}{3}-\frac{1}{3}\Rightarrow \frac{x}{3}-\frac{x}{3}=-\frac{1}{3}-\frac{1}{2}\Rightarrow \frac{3x-2x}{6}=\frac{-2-3}{6}$   [Group like term each side and solve]

$\Rightarrow \frac{x}{6}=\frac{-5}{6}\Rightarrow x=\frac{-5}{6}\times 6=-5\Rightarrow =-5$ is the solution or root of the given equation.

Find the value of x which satisfies the equation $\frac{{{x}^{2}}+5}{2-{{x}^{2}}}=\frac{-3}{2}.$

(a)$x=5$

(b) $x=\pm 5$

(c)$x=\pm 4$

(d) $x=9$

(e) None of these

Explanation

Putting ${{x}^{2}}=y$in the given equation we get $\frac{y+5}{2-y}=\frac{-3}{2}.$

By cross multiplication, $2(y+5)=-3(2-y)$

Or, $2y+10=-6+3y\Rightarrow 2y-3y=-6-10$

Or, $-y=-16,y=16$or $\text{ }or,\text{ }x=\pm 4$

Find the solution of the equation $\frac{2x-1}{3x+5}=5.$

(a)$~x=-3$

(b)$~x=-5$

(c)$x=-2$

(d) $x=-9$

(e) None of these

Explanation

We have $\frac{2x-1}{3x+5}=5$ $\frac{2x-1}{3x+5}\times (3x+5)=5x(3x+5)$(Multiplying both sides by $~(3x+5)$

$\Rightarrow 2x-1=15x+25$

$\Rightarrow 2x-1=15x+25$[Transposing 15x, to L.H.S. & (-1) to R.H.S.]

$\Rightarrow -13=26,$or $x=\frac{26}{-13}=-2$

Which one of the following options is the solution of the equation $\frac{4x-3}{2x+3}=\frac{5}{7}?$

(a) 2

(b) 4

(c) 10

(d) 9

(e) None of these

Explanation

$\frac{4x-3}{2x+3}=\frac{5}{7}$Multiplying both sides of the above equation by $(2x+3)$

$\frac{4x-3}{2x+3}\times (2x+3)=\frac{5}{7}\times (2x+3)$

$\Rightarrow 4x-3=\frac{5}{7}(2x+3)\Rightarrow 4x-3$

$=\frac{10x+15}{7}\Rightarrow 7(4x-3)=10x+15$

$\Rightarrow 28x-21=10x+15\Rightarrow 28x-10x$

$=15+21\Rightarrow 18x=36\Rightarrow x=\frac{36}{18}=2$

If $\frac{(z+2)}{4}+\frac{z-3}{5}=\frac{5z-4}{6},$ then value of z which satisfies the given equation is:

(a) Any natural number

(b) A natural number lies between 7 and 9

(c) A natural number less than 8

(d) It is not a natural number

(e) None of these

If $\frac{z+6}{4}+\frac{z-3}{5}=\frac{5z-4}{6}$ then the value of z is

(a) 10

(b) $-10$

(c) Cannot be determined

(d) 5

(e) None of these

If $\frac{x-3}{x+4}=\frac{x+1}{x-2},$ then which of the following value satisfies it?

(a) 1

(b) $\frac{2}{19}$

(c) $\frac{2}{12}$

(d) $\frac{2}{10}$