SOLVING EQUATIONS WITH VARIABLES ON BOTH SIDES

Let us look at the common steps involved in solving equations with variable on both sides.  

Step 1 :

Get rid of the variable on one of the sides of the equations using the mathematical operations like addition or subtraction.

When we get rid of the variable on one of the sides of the equation, always we have to take care that the variable on the other side must be positive. 

Step 2 :

Simplify the expressions on each side of the equal sign.

Step 3 :

Use the operations addition, subtraction, multiplication and division to solve for the variable. 

Note : 

When we solve equations with variable on both sides, sometimes the variable may vanish at the last step.

That is, there will be no variable. 

In such a case, if the result at the last step is true, then the equation has infinitely many solutions. 

If the result at the last step is false, then the equation has no solution.

Example 1 :

Solve for x : 

-7x - 3x + 2  =  -8x - 8

Solution :

-7x - 3x + 2  =  -8x - 8

Simplify.

-10x + 2  =  -8x - 8

Add 10x to each side. 

2  =  2x - 8

Add 8 to each side.

10  =  2x

Divide each side by 2.

5  =  x

Example 2 :

Solve for k :

5(k - 3) - 7(6 - k)  =  24 - 3(8 - k) - 3

Solution :

5(k - 3) - 7(6 - k)  =  24 - 3(8 - k) - 3

Use distributive property. 

5k - 15 - 42 + 7k  =  24 - 24 + 3k - 3

Simplify. 

12k - 57  =  3k - 3

Subtract 3k from each side. 

9k - 57  =  -3

Add 57 to each side.

9k  =  54

Divide each side by 9.

k  =  6

Example 3 :

Solve for x : 

(x + 15)(x - 3) - (x2 - 6x + 9)  =  30 - 15(x - 1)

Solution :

(x + 15)(x - 3) - (x2 - 6x + 9)  =  30 - 15(x - 1)

Simplify. 

x2 + 12x - 45 - x2 + 6x - 9  =  30 - 15x + 15

18x - 54  =  -15x + 45

Add 15x to each side. 

33x - 54  =  45

Add 54 to each side.

33x  =  99

Divide each side by 33.

x  =  3

Example 4 :

Solve for x :  

(4x/5) - (7/4)  =  (x/5) + (x/4)

Solution :

(4x/5) - (7/4)  =  (x/5) + (x/4)

The least common multiple of the denominators in the equation is 4 × 5  =  20 and we proceed as follows :

20[4x/5 - 7/4]  =  20[x/5 + x/4]

20(4x/5) - 20(7/4)  =  20(x/5) + 20(x/4)

16x - 35  =  4x + 5x

16x - 35  =  9x

Subtract 9x from each side. 

7x - 35  =  0

Add 35 to each side.

7x  =  35

Divide each side by 7.

x  =  5

Example 5 :

Solve for x : 

5x - (4x - 7)(3x - 5)  =  6 - 3(4x - 9)(x - 1)

Solution :

5x - (4x - 7)(3x - 5)  =  6 - 3(4x - 9)(x - 1)

Simplify. 

5x - (12x2 - 41x + 35)  =  6 - 3(4x2 - 13x + 9)

5x - 12x2 + 41x - 35  =  6 - 12x2 + 39x - 27

- 12x2 + 46x - 35  =   -12x2 + 39x - 21

Add 12x2 to each side. 

46x - 35  =  39x - 21

Subtract 39x from each side. 

7x - 35  =  -21

Add 35 to each side.

7x  =  14

Divide each side by 7.

x  =  2

Example 6 :

Solve for x :  

(x - 2)/2 + (x + 10)/9  =  5

Solution :

(x - 2)/2 + (x + 10)/9  =  5

The least common multiple of the denominators in the equation is 2 × 9  =  18 and we proceed as follows :

18[(x - 2)/2 + (x + 10)/9]  =  18(5)

18(x - 2)/2 + 18(x + 10)/9  =  90

9(x - 2) + 2(x + 10)  =  90

9x - 18 + 2x + 20  =  90

11x + 2  =  90

Subtract 2 from each side. 

11x  =  88

Divide each side by 11. 

x  =  8

Example 7 :

Solve the following equation : 

(1/2)(8y - 6)  =  5y - (y + 3)

Solution :

(1/2)(8y - 6)  =  5y - (y + 3)

Simplify both sides. 

4y - 3  =  5y - y - 3

4y - 3  =  4y - 3  

Subtract 4y from each side. 

-3  =  -3

The above result is true. Because the result we get at the last step is true, the given equation has infinitely has many solutions. 

Example 8 :

Solve the following equation : 

2(1 - x) + 5x  =  3(x + 1)

Solution :

2(1 - x) + 5x  =  3(x + 1)

Simplify both sides. 

2 - 2x + 5x  =  3x + 3

2 + 3x  =  3x + 3

Subtract 3x from each side. 

2  =  3

The above result is false. Because 2 is not equal to 3. Because the result we get at the last step is false, the given equation has no solution.

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