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How to Solve Math Equations

                       Math Equations are those where two algebraic expressions involving a  one or more variables are equated. If the equation involves only one variable, it is called a linear equation in one variable. We need to have only one equation to solve the equation for the variable. IF the equation involves two or more variables, we need to have as many linear equations ( also called simultaneous linear equations ) as there are variables. The solutions to these equations give the point of intersection of the lines or the planes as the case may be. We can also find the points of intersection of the curves by solving the equations of the curves. In this section let us see the method of solving some of the math equations.

Help Solving Math Equations :

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Solving linear equations of one variable:

1. Two Step Method: In this we first eliminate the constant by adding the additive identity of the constant. Then we multiply both sides by the reciprocal of the coefficient of the variable to find the value of the variable, which is the solution to the given equation.

Example : 3x + 5 = 29

Solution: 3x +  5 = 29
 =>      3x + 5 - 5 = 29 - 5 [ adding both sides the additive inverse of 5 which is -5 ]
 =>                 3x = 24      [ multiplying both sides by the reciprocal of 3 which is 1/3 ]
 =>         $\frac{1}{3}$ ( 3x ) = $\frac{1}{3}$ ( 24 )
 =>                  x = 8
Therefore the solution to the above equation is x = { 8 }

2. Five Steps Method: In this we follow the following steps.
Step 1: Remove the parenthesis on both sides of the equation by multiplying by the appropriate integer before the parenthesis.
Step 2: Combine the like terms on each side of the equation.
Step 3: Add or subtract the appropriate terms on both sides such that the terms containing the variable is on the left side of the equation
Step 4: Add or subtract the appropriate constants on both sides of the equation such that and the constant terms lies on the right side of the equation.
Step 5: Divide both sides of the equation by the coefficient of x, which gives the correct solution of the equation.

Example : Solve : 2 ( 4x - 17 ) = 3 ( x - 3 )

Solution: We have, 2 ( 4x - 17 ) = 3 ( x - 3 )
                   =>             8x - 34 = 3x - 9  [ multiplying by the integer outside the parenthesis ]
                   =>       8x - 3x - 34 =  3x - 9  - 3x [ subtracting 3x from both sides of the equation ]
                   =>             5x  - 34 =  - 9
                   =>      5x - 34 + 34 = -9 + 34 [ adding 34 on both sides of the equation ]
                   =>                     5x = 25
                   =>                   $\frac{5x}{5}$ = $\frac{25}{5}$ [ dividing by 5 on both sides of the equation ]
                  =>                       x = 5

Solving Linear Equations of 2 variables :

The linear equations of two variables can be solved if we have at least two equations containing the two variables.The pair equations containing the variables is also called as simultaneous equations of two variables. Because the pair of equations have a common solution which satisfy both the equations simultaneously.
Example : Check if ( 2, 3 ) is the solution of the equations, x + y = 5 and 2x - y = 1
Solution:
Let us plug in the values x = 2 and y = 3 in the two equations,

                                         x + y = 5
                                    => 2 + 3 = 5
                                    =>      5 = 5
                                    => the point ( 2, 3 ) is the solution of the equation x + y = 1.
              Let us plug in x = 2 and y = 3 in the equation, 2x - y = 1
                                    => 2 ( 2) - 3 = 1
                                    => 4 - 3 = 1
                                    => 1 = 1
                                    => ( 2, 3) is the solution of the equation 2x - y = 1
Therefore ( 2, 3 ) is the common solution of the simultaneous equations 2x - y = 1 and x + y = 5
          

Linear Equation Graph

We observe from the above graph that the two lines intersect at the point ( 2, 3 ).
Solving second degree equations: The graph of  second degree equations is a curve. By solving the second degree equations we get the point of intersection of the two curves.
Example : Solution to the second degree equations x2 + y2 = 25 and 16 x  = 3y2

Solution : We have the equations, x2 + y2 = 25 --------------------------( 1 )
                      and                            16 x = 3 y2  ------------------------(2)
Multiplying Equation (1) by 3, we get,
                                                         3 x2 + 3 y2 = 3 x 25 = 75
                               = >                    3 x2 + 16 x  =  75
                               =>              3 x2 + 16 x  -  75 = 0
                               =>        3 x2 + 25 x - 9 x - 75  = 0
                               => x ( 3x + 25 ) - 3 ( 3x + 25 ) = 0
                               =>            ( x - 3 ) ( 3 x + 25 ) = 0
                               =>                                       x = 3 or x = - 25 /3
                   Value of x cannot be negative, since the curve opens towards right.
Therefore x = 3. substituting x = 3 in            16 x = 3 y2
                                                            16 ( 3 ) = 3 y2
                                             =>               3 y2 = 48
                                             =>                  y2 = 48/3 = 16
                                             =>                    y = $\pm$ 4
The solutions to the above equations will be ( 3, 4  ) and ( 3, - 4 )

The points of intersection will be ( 3, 4 ) and (3, - 4 )     
    
When we draw the graph using graphing calculator we get the graphs as shown below.
Solution
We observe from the above graph that the two curves intersect at the points ( 3, 4 ) and ( 3, -4 )

Practice Questions:

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1. Solve the equation 4x + 10 = - 34 by two steps method.
2. Solve 2 ( x + 4 ) = 5 ( 3 - x ), by 5 steps method.
3. Solve the simultaneous linear equations, y = 3x - 5 and 2x + 3y = 18 by substitution method.
4. Solve the second degree equations, $\left ( \frac{x^{2}}{9} \right )$ + $\left ( \frac{y^{2}}{16} \right )$ = 1
and $\left ( \frac{y^{2}}{9} \right )$ + $\left ( \frac{x^{2}}{16} \right)$ = 1

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