8th Class Mathematics Linear Equations in One Variable Linear Equation in One Variable

         Introduction of Linear Equation

An algebraic equation is an equality involving one or more variables. It has equality sign in between. The expression on the left is called LHS and the expression on the right is called RHS. An expression in one variable having degree one is called linear equation in one variable. In a linear equation the value of the expression on LHS and RHS are all equal. The value of the variable which satisfies the given expression is called solution of the linear equation. We can find the solution of the linear equation either by hit and trial method or by solving the given equation for the required variables.  

            How to solve Linear Equation in One Variable

There are some set of rules to solve linear equation in one variable:

• We can add the same number on both sides of the equation.
• We can subtract the same number from both sides of the equation.
• We can multiply both sides of the equation by the same number.
• We can divide both sides of the equation by some none zero numbers.
• If \[\frac{ax+by}{cx+dy}=\frac{p}{q}\] then \[q(ax+b)=p(cx+d)\].
• This process is called cross multiplication.
• Any term of the equation can be transferred to any side of the equation by changing the sign of the term. This is called transposition.  

            Properties of Equality

Let us consider an algebraic expression as A, B and C.  

Property 1:

Addition Property of Equality.                       

If \[\text{A}=\text{B}\] then \[\text{A}+\text{C}=\text{B}+\text{C}\]  

Property 2:

Subtraction Property of Equality.                       

If\[\text{A}=\text{B}\] then \[\text{A}-\text{C}=\text{B}-\text{C}\]  

Property 3:

Multiplication Property of Equality.                       

If\[\text{A}=\text{B}\] and \[C\ne 0\] then CA = CB.  

Property 4:     

Division Property of Equality.                      

If\[\text{A}=\text{B}\] and \[C\ne 0\] then \[\frac{A}{C}=\frac{B}{C}\]

Note: Multiplying or dividing both sides of an equation by zero is carefully avoided. Dividing by zero is undefined and multiplying both sides by zero will result in an equation 0=0.  

• A linear equation can be used to solve the word problems related to day to day situations.
• A linear equation cannot have more than one solution.
• A linear equation in two variables always represents a straight line on the graph.
• The solution of the linear equation is that number which satisfies the given equations.
• There can be infinite number of solutions of the linear equation of two variables.  

• An algebraic expression is an equality involving one or more variables.
• The solution of the linear equation is the value of the variable which is obtained by solving the given equation.
• A linear equation may have a rational solution.
• In a linear equation variable can also be transposed.
• Linear equation can be used for the purpose of problem solving.    

  Find the solution of the given linear equation \[\frac{6m+7}{3m+2}=\frac{4m+5}{2m+3}\].

(a) \[-\frac{11}{9}\]                                        

(b) \[\frac{11}{9}\]        

(c) \[\frac{9}{11}\]                                                          

(d) \[-\frac{9}{11}\]

(e) None of these  

Answer: (a)

Exploration:                

\[=\frac{6m+7}{3m+2}=\frac{4m+5}{2m+3}\]

\[\Rightarrow \]\[(6m+7)(2m+3)=(4m+5)(3m+2)\]          

\[\Rightarrow \]\[12{{m}^{2}}+18m+14m+21=21{{m}^{2}}+8m+15m+10\]

\[\Rightarrow \]\[32m+21=23m+10\]

\[\Rightarrow \]\[m=-\frac{11}{9}\]  

  Find the solution of \[\frac{4y+1}{3}+\frac{2y-1}{2}-\frac{3y-7}{5}=6\]

(a) 1                                                      

(b) \[-2\frac{11}{4}\]        

(c) \[-\frac{11}{4}\]                                        

(d) \[2\frac{3}{4}\]

(e) None of these  

Answer: (d)

Explanatio:

We have,

\[\Rightarrow \]\[\frac{4y+1}{3}+\frac{2y-1}{2}-\frac{3y-7}{5}=6\]

\[\Rightarrow \]\[\frac{40y+10+30y-15-18y+42}{30}=6\]

\[\Rightarrow \]\[52y+37=180\]

\[\Rightarrow \]\[52y=180-37\]

\[\Rightarrow \]\[y=\frac{143}{52}=\frac{11}{4}\]  

  Find the value of n such that \[\frac{2}{3}(4n-1)\]\[-(2n-\frac{1+n}{3})=\frac{1}{3}n+\frac{4}{3}\].

(a) \[-\frac{5}{4}\]                                          

(b) \[\frac{5}{4}\]

(c) \[-\frac{5}{2}\]                                           

(d) \[\frac{5}{2}\]

(e) None of these  

Answer: (d)  

  Solve the equation \[\frac{0.5(z-0.4)}{3.5}-\frac{0.6(z-2.7)}{4.2}\]\[=z+6.1\]

(a) \[-\frac{202}{35}\]                                                   

(b) \[\frac{202}{35}\]

(c) \[\frac{35}{202}\]                                     

(d) \[-\frac{35}{202}\]

(e) None of these  

Answer: (a)

Explanation:

\[\Rightarrow \]\[\frac{5(z-0.4)}{35}-\frac{6(z-2.7)}{42}=z+6.1\]

\[\Rightarrow \]\[\frac{30z-12-30z+81}{210}=z+6.1\]

\[\Rightarrow \]\[\frac{69}{210}=z+6.1\]

\[\Rightarrow \]\[z=-\frac{202}{35}\]