Hints for solving the problems - Apply 2.2

Problem No. 2  –   Is   y  =  -5  a solution of  y  +  3  =  -2 ?

Substitute the value of y for y in the equation, perform the operations and if there is a true statement, then the value of y is a solution. If there is a false statement then it is not a value for y.

                     -5 + 3 = -2

                          -2 = -2       Yes -5 is a solution for the equation.

Problem No. 8  –  Solve for x:   15  -  x  =  16

To solve an equation first use the addition or subtraction property (add or subtract the same term from both sides of the equation). Then use the multiplication or division property (multiply or divide the numerical coefficient of the variable by both sides of the equation). The idea is to isolate the variable (usually on the left side) and the constants (numbers) on the other and solve for the variable.

                  15  -  x  =  8

                 -15          -15         Subtract 15 from both sides.

                        - x   =  -7

                        - 1       -1         Divide both sides by -1

                           x    =   7        The answer

Problem No. 20  –  Solve for h:    12  +  5h  =  -38

                                       -12                   -12   Subtract 12 from both sides of the equation.

                                                   5h   =    -50

                                                    5             5     Divide both sides of the equation by 5.

                                                      h   =   -10     The answer

Problem No. 30  –  Solve for y:   12y   -   13   =   7y   +   12

Notice in this problem there is a term with a variable and a constant on both sides of the equation. Terms must be moved. To be consistent, move the constant term on the left side of the equation to the right side of the equation and the term with the variable on the right side of the equation to the left side of the equation.

                   12y   -   13   =   7y   +   12

                                 13                    13       Add 13 to both sides of the equation.

                   12y              =    7y   +   25

                   -7y                    -7y                 Add -7y to both sides of the equation.

                           5y        =                25

                            5                             5      Divide both sides of the equation by 5

                                y   =   5                       Is the answer.

Problem No. 34   –   Solve for x:   4(x   +   3)   =   -5(3x   -  10)

                                4(x  +  3)  =  -5(3x  -  10)

                                  4x  + 12  =   -15x  +  50      Using the distributive property.

                                         - 12                  - 12      Subtracting 12 from both sides.

                                  4x           =    -15x  +  38

                                15x           =      15x              Adding 15x to both sides.

                                            19x = 38

                                            19      19                   Dividing by 19

                                               x   =   2          Is the answer

Problem No. 42  –  Solve for m:    

Notice this problem has fractions. First step is to multiply the LCD (Least Common Denominator) by every term in the problem. The LCD (10) will be multiplied by the 12, (-3/10)m and (7/10)(m + 20).

              120 - 3m = 7(m + 20)

                                                             120 - 3m = 7m + 140

                                                            -120                     -120

                                                                    - 3m =  7m  +  20

                                                                    - 7m =  -7m

                                                                    -10m =          20

                                                                     -10               -10

                                                                   m = -2 is the answer.

Problem No. 46  –  Solve for p:   

The usual way to solve this problem is to multiply all terms on both sides of the equation by the LCD (2). But if you notice that you can multiply the ½ by every term in the parenthesis their will be a whole number left. DO IT.

                 3p  +  6  =  3p  +  6

                                                    - 6               - 6

                                            3p          =   3p

                                           -3p              -3p

A true statement 0 = 0   The answer is all Real Numbers.

When you solve a problem and the variables are eliminated from the problem and there are only constants left, look at the answer. If it is a true statement, such as: 0 = 0, then the answer is all Real Numbers. But if the statement is not true, such as: 0 = 5, then the answer is no solution, or null set (Ø).

When you multiplied the ½ by both terms on the parenthesis and got

3p + 6 = 3p + 6, notice both sides of the equation have the same terms.

So the answer would be all Real Numbers or any number. If the variable term was the same and the constants different then the answer would be no solution.

Problem No. 48  –  Solve for y:   

In this problem I would remove the parenthesis first, because the fraction is inside of the parenthesis.

                    Multiply by 2 (LCD)

                 30 + y     =  16  +   y

                -30               -30

                         y      = -14   +  y

                        -y      =            -y

A false statement 0 = -14   The answer is no solution.

Problem No. 54  –  Solve for c    5b   -   2c   =   10

These equations are called literal equations. You will solve for one variable in terms of another variable or variables. In this problem you must isolate the c and solve for it in terms of b.

                      5b   -   2c   =   10

                     -5b                    -5b      Subtract 5b from both sides of equation.

                                 -2c   =  10  -  5b

                                 -2               -2               Divide by -2

                                  c  =  10  -  5b    or   c  =  5b  -  10

                                                -2                              2

Either of the above is the correct answer. Some instructors and textbooks will not let you leave a negative (-) number in the denominator, so you have to multiply both the numerator and denominator by -1 to eliminate the negative sign in the denominator.

Problem No. 56   –   Solve for t:   

     Multiply every term by the LCD (2) first.

               t + 6v = 10

                                          t  =  10  -  6v the answer