7.3 Simple Algebraic Equations and Word Problems

An algebraic equation is a mathematical sentence expressing equality between two algebraic expressions (or an algebraic expression and a number).

When two expressions are joined by an equal (=) sign, it indicates that the expression to the left of the equal sign is identical in value to the expression to the right of the equal sign.

For example, when two algebraic expressions, such as [latex]5x + 7[/latex] and [latex]x + 19[/latex], are equal, the two expressions are joined by an equal (=) sign and the equation is written as:

[latex]5x + 7 = x + 19[/latex]

‘Left side’ (LS) = ‘Right side’ (RS)

The solution to the equation is the value of the variable that makes the left side (LS) evaluate to the same number as the right side (RS).

Note: You need an equation to solve for an unknown variable – you cannot solve for a variable in an algebraic expression that is not part of an equation.

• If you have an expression, it needs to be simplified.
• If you have an equation, it needs to be solved.

In algebra, there are a variety of equations. In this section, we will learn one equation category, linear equations with one variable.

Examples of linear equations with one variable are:

[latex]2x = 8[/latex],   [latex]3x + 5 = 14[/latex],   [latex]5x + 7 = x + 19[/latex]

An equation is either true or false depending on the value of the variable.

For example, consider the equation [latex]2x = 8[/latex]:

• If [latex]x = 4[/latex],   LS = 2(4) = 8, RS = 8; therefore, the equation is true.
• If [latex]x = 3[/latex],   LS = 2(3) = 6, RS = 8; therefore, the equation is false.

Equations may be classified into the following three types:

1. Conditional equation: these equations are only true when the variable has a specific value. For example, [latex]2x = 8[/latex] is a conditional equation, true if and only if [latex]x = 4[/latex].
2. Identity: these equations are true for any value for the variable. For example, [latex]2x + 10 = 2(x + 5)[/latex] is an identity, true for any value of [latex]x[/latex].
3. Contradiction: these equations are not true for any value of the variable. For example, [latex]x + 5 = x + 4[/latex] is a contradiction, not true for any value of [latex]x[/latex].

Equivalent Equations

Equations with the same solutions are called equivalent equations.

For example, [latex]2x + 5 = 9[/latex] and [latex]2x = 4[/latex] are equivalent equations because the solution [latex]x = 2[/latex] satisfies each equation.

Similarly, [latex]3x - 4 = 5[/latex], [latex]2x = x + 3[/latex], and [latex]x + 1 = 4[/latex] are equivalent equations because the solution [latex]x = 3[/latex] satisfies each equation.

Properties of Equality

If [latex]a = b[/latex], then,

These properties are used to solve equations.

Equations with Fractional Coefficients

If an equation contains fractional coefficients, then the fractional coefficients can be changed to whole numbers by multiplying both sides of the equation by the least common denominator (LCD) of all the fractions, using the Multiplication Property.

For example,

[latex]\displaystyle{\frac{2}{3}x = \frac{5}{2} + 4}[/latex]
Since the LCD of the denominators 3 and 2 is 6, multiply both sides of the equation by 6.

[latex]\displaystyle{6\left(\frac{2}{3}x\right) = 6\left(\frac{5}{2} + 4\right)}[/latex]
This is the same as multiplying each term by the LCD of 6.

[latex]\displaystyle{6\left(\frac{2}{3}x\right) = 6\left(\frac{5}{2}\right) + 6(4)}[/latex]
Simplifying, [latex]4x = 15 + 24[/latex]
Now, the equation has only whole number coefficients [latex]4x = 39[/latex].

Equations with Decimal Coefficients

If an equation contains decimal coefficients, then the decimal coefficients can be changed to whole numbers by multiplying both sides of the equation by an appropriate power of 10, using the Multiplication Property.

For example,

[latex]\underline{1.25}x = \underline{0.2} + 4[/latex]
Since there is at most 2 decimal places in any of the coefficients or constants, multiply both sides of the equation by [latex]10^2 = 100[/latex].

[latex]100(1.25x) = 100(0.2 + 4)[/latex]
This is the same as multiplying each term by 100. [latex]100(1.25x) = 100(0.2) + 100(4)[/latex]
Simplifying, [latex]125x = 20 + 400[/latex]
Now, the equation has only whole number coefficients [latex]125x = 420[/latex].

Steps to Solve Algebraic Equations with One Variable

Step 1:

If the equation contains fraction and/or decimal coefficients, it is possible to work with them as they are – in that case, proceed to Step 2. Alternatively, as explained earlier, the equation may be rewritten in whole numbers to make calculations and rearrangements easier.

Step 2:

If present, expand and clear brackets in the equation by following the order of arithmetic operations (BEDMAS).

Step 3:

Use the addition and subtraction properties to collect and group all variable terms on the left side of the equation and all constants on the right side. Then, simplify both sides.

Note: If it is more convenient to gather all the variable terms on the right side and the constants on the left side, you may do so, and then use the symmetric property and switch the sides of the equation to bring the variables over to the left side and the constants to the right side.

Step 4:

Use the division and multiplication properties to ensure that the coefficient of the variable is +1.

Step 5:

After completing Step 4, there should be a single variable with a coefficient of +1 on the left side and a single constant term on the right side – that constant term is the solution to the equation.

Step 6:

Verify the answer by substituting the solution from Step 5 back into the original problem.

Step 7:

State the answer.

Solution

1. [latex]x - 11 = 4[/latex]
   Adding 11 to both sides, [latex]x - 11 + 11 = 4 + 11[/latex]
   [latex]x = 15[/latex]
   Verify by substituting [latex]x = 15[/latex]:
   LS [latex]= x - 11 = 15 - 11 = 4[/latex]     RS [latex]= 4[/latex]
   LS = RS
   Therefore, the solution is [latex]x = 15[/latex].
2. [latex]8 + x = 20[/latex]
   Subtracting 8 from both sides, [latex]8 - 8 + x = 20 - 8[/latex]
   [latex]x = 12[/latex]
   Verify by substituting [latex]x = 12[/latex]:
   LS [latex]= 8 + x = 8 + 12 = 20[/latex]   RS [latex]= 20[/latex]
   LS = RS
   Therefore, the solution is [latex]x = 12[/latex].

Solution

1. [latex]5x = 20[/latex]
   Dividing both sides by 5, [latex]\displaystyle{\frac{5x}{5} = \frac{20}{5}}[/latex]
   [latex]x = 4[/latex]
   Verify by substituting [latex]x = 4[/latex]:
   LS [latex]= 5x = 5(4) = 20[/latex]   RS [latex]= 20[/latex]
   LS = RS
   Therefore, the solution is [latex]x = 4[/latex].
2. [latex]\displaystyle{\frac{3}{8}x = 12}[/latex]
   Multiplying both sides by [latex]\displaystyle{\frac{8}{3}}[/latex] (the reciprocal of [latex]\displaystyle{\frac{3}{8}}[/latex]), [latex]\displaystyle{\left(\frac{8}{3}\right) \cdot \frac{3}{8}x = \left(\frac{8}{3}\right) \cdot 12}[/latex]
   [latex]x = 8 \times 4[/latex]
   [latex]x = 32[/latex]   or  [latex]\displaystyle{\frac{3}{8}x = 12}[/latex]
   Multiplying both sides by 5, [latex]\displaystyle{(8) \cdot \frac{3}{8}x = (8) \cdot 12}[/latex]
   [latex]3x = 96[/latex]Dividing both sides by 3, [latex]\displaystyle{\frac{3x}{3} = \cdot \frac{96}{3}}[/latex][latex]x = 32[/latex]Verify by substituting [latex]x = 32[/latex]:LS [latex]\displaystyle{= \frac{3}{8}x = \frac{3}{8} \times 32 = 12}[/latex]RS [latex]= 12[/latex]LS = RSTherefore, the solution is [latex]x = 32[/latex].

Solution

1. [latex]3x - 8 = 12 - 2x[/latex]
   Adding 2x to both sides,[latex]3x + 2x - 8 = 12 - 2x + 2x[/latex]
   [latex]5x - 8 = 12[/latex]
   Adding 8 to both sides,[latex]5x - 8 + 8 = 12 + 8[/latex]
   [latex]5x = 20[/latex]
   Dividing both sides by 5, [latex]\displaystyle{\frac{5x}{5} = \frac{20}{5}}[/latex]
   [latex]x = 4[/latex]
   Verify by substituting [latex]x = 4[/latex]:
   LS [latex]= 3x - 8 = 3(4) - 8 = 12 - 8 = 4[/latex]RS [latex]= 12 - 2x = 12 - 2(4) = 12 - 8 = 4[/latex]LS = RSTherefore, the solution is [latex]x = 4[/latex].
2. [latex]15 + 6x - 4 = 3x + 31 - x[/latex]
   Combining like terms (LS: [latex]15 - 4 = 11[/latex], and RS: [latex]3x - x = 2x[/latex]), [latex]11 + 6x = 2x + 31[/latex]   Subtracting [latex]2x[/latex] from both sides, [latex]11 + 6x - 2x = 2x - 2x + 31[/latex]
   [latex]11 + 4x = 31[/latex]
   Subtracting 11 from both sides, [latex]11 - 11 + 4x = 31 - 11[/latex]
   [latex]4x = 20[/latex]
   Dividing both sides by 4, [latex]\displaystyle{\frac{4x}{4} = \frac{20}{4}}[/latex]
   [latex]x = 5[/latex]
   Verify by substituting [latex]x = 5[/latex] back into the original equation:LS [latex]= 15 + 6x - 4 = 15 + 6(5) - 4 = 15 + 30 - 4 = 41[/latex]RS [latex]= 3x + 31 - x = 3(5) + 31 - 5 = 15 + 31 - 5 = 41[/latex]LS = RSTherefore, the solution is [latex]x = 5[/latex].

Solution

[latex]\displaystyle{\frac{x}{3} - \frac{1}{12} = \frac{1}{6} + \frac{x}{4}}[/latex]
LCD of 3, 4, 6, and 12 is 12.

Multiplying each term by 12,

[latex]\displaystyle{12\left(\frac{x}{3}\right) - 12\left(\frac{1}{12}\right) = 12\left(\frac{1}{6}\right) + 12\left(\frac{x}{4}\right)}[/latex]

[latex]4x - 1 = 2 + 3x[/latex]

Subtracting [latex]3x[/latex] from both sides,

[latex]4x - 3x - 1 = 2 + 3x - 3x[/latex]

[latex]x - 1 = 2[/latex]

Adding 1 to both sides,

[latex]x - 1 + 1 = 2 + 1[/latex]

[latex]x = 3[/latex]

Verify by substituting [latex]x = 3[/latex] back into the original equation:

LS [latex]\displaystyle{= \frac{x}{3} - \frac{1}{12} = \frac{3}{3} - \frac{1}{12} = \frac{12}{12} - \frac{1}{12} = \frac{11}{12}}[/latex]

RS [latex]\displaystyle{= \frac{1}{6} + \frac{x}{4} = \frac{1}{6} + \frac{3}{4} = \frac{2}{12} + \frac{9}{12} = \frac{11}{12}}[/latex]

LS = RS

Therefore, the solution is [latex]x = 3[/latex].

Solution

[latex]0.15x + 1.2 = 0.4x - 0.8[/latex]
Greatest number of decimal places is 2 (i.e., hundredths).

Multiplying all the terms by [latex]10^2 = 100[/latex],

[latex]100(0.15x) + 100(1.2) = 100(0.4x) - 100(0.8)[/latex]

[latex]15x + 120 = 40x - 80[/latex]

Interchanging the LS and RS using the Symmetric Property to have the larger x term on the LS,

[latex]40x - 80 = 15x + 120[/latex]

Subtracting [latex]15x[/latex] from both sides,

[latex]40x - 15x - 80 = 15x - 15x + 120[/latex]

[latex]25x - 80 = 120[/latex]

Adding 80 to both sides,

[latex]25x - 80 + 80 = 120 + 80[/latex]

[latex]25x = 200[/latex]

Dividing both sides by 25,

[latex]\displaystyle{\frac{25x}{25} = \frac{200}{25}}[/latex]

[latex]x = 8[/latex]

Verify by substituting [latex]x = 8[/latex] back into the original equation:

LS [latex]= 0.15x + 1.2 = 0.15(8) + 1.2 = 1.2 + 1.2 = 2.4[/latex]

RS [latex]= 0.4x - 0.8 = 0.4(8) - 0.8 = 3.2 - 0.8 = 2.4[/latex]

LS = RS

Therefore, the solution is [latex]x = 8[/latex].

Note: For the rest of the examples in this section, we will not show the verification by substitution step.

Solution

1. [latex]8x + 7 - 3x = -6x - 15 + x[/latex]
   Grouping like terms on both sides, [latex]8x - 3x + 7 = -6x + x - 15[/latex]
   [latex]5x + 7 = -5x - 15[/latex]
   Adding 5 to both sides, [latex]5x + 5x + 7 = -5x + 5x - 15[/latex]
   [latex]10x + 7 = -15[/latex]
   Subtracting 7 from both sides, [latex]10x + 7 - 7 = -15 - 7[/latex]
   [latex]10x = -22[/latex]
   Dividing both sides by 10, [latex]\displaystyle{\frac{10x}{10} = -\frac{22}{10}}[/latex]
   [latex]\displaystyle{x = -\frac{11}{5} = -2\frac{1}{5}}[/latex]
2. [latex]2(3x - 7) = 28 - 3(x + 1)[/latex]
   Expanding both sides,[latex]6x - 14 = 28 - 3x - 3[/latex]   Grouping like terms,[latex]6x - 14 = 28 - 3 - 3x[/latex]
   [latex]6x - 14 = 25 - 3x[/latex]
   Adding [latex]3x[/latex] to both sides,[latex]6x + 3x - 14 = 25 - 3x + 3x[/latex]
   [latex]9x - 14 = 25[/latex]
   Adding 14 to both sides, [latex]9x - 14 + 14 = 25 + 14[/latex]
   [latex]9x = 39[/latex]
   Dividing both sides by 9,[latex]\displaystyle{\frac{9x}{9} = \frac{39}{9}}[/latex]
   [latex]\displaystyle{x = \frac{13}{3} = 4\frac{1}{3}}[/latex]
3. [latex]\displaystyle{\frac{1}{4}(x + \frac{2}{3}) = \frac{1}{2}(x - 3) + x}[/latex]
   Expanding both sides, [latex]\displaystyle{\frac{1}{4}x + \frac{1}{6} = \frac{1}{2}x - \frac{3}{2} + x}[/latex]
   Multiplying each term by the LCD 12, [latex]\displaystyle{12\left(\frac{1}{4}x\right) + 12\left(\frac{1}{6}\right) = 12\left(\frac{1}{2}x\right) - 12\left(\frac{3}{2}\right) + 12(x)}[/latex]
   [latex]3x + 2 = 6x - 18 + 12x[/latex]
   Grouping like terms,[latex]3x + 2 = 6x + 12x - 18[/latex]
   [latex]3x + 2 = 18x - 18[/latex]
   Interchanging the LS and RS using the Symmetric Property to have the larger x term on the LS, [latex]18x 18 = 3x + 2[/latex]   Subtracting [latex]3x[/latex] from both sides, [latex]18x - 3x - 18 = 3x - 3x + 2[/latex]
   [latex]15x - 18 = 2[/latex]
   Adding 18 to both sides, [latex]15x - 18 + 18 = 2 + 18[/latex]
   [latex]15x = 20[/latex]
   Dividing both sides by 15, [latex]\displaystyle{\frac{15x}{15} = \frac{20}{15}}[/latex][latex]\displaystyle{x = \frac{4}{3} = 1\frac{1}{3}}[/latex]
4. [latex]0.45(2x + 3) - 2.55 = 0.6(3x - 5)[/latex]
   Expanding both sides, [latex]0.90x + 1.35 - 2.55 = 1.8x - 3.0[/latex]
   Greatest number of decimal places is 2 (i.e., hundredths).
   Multiplying all the terms by [latex]10^2 = 100[/latex],   [latex]100(0.90x) + 100(1.35) - 100(2.55) = 100(1.8x) - 100(3.0)[/latex]
   [latex]90x + 135 - 255 = 180x - 300[/latex]
   Grouping like terms,[latex]90x 120 = 180x 300[/latex]
   Interchanging the LS and RS using the Symmetric Property to have the larger x term on the LS,   [latex]180x - 300 = 90x - 120[/latex]
   Subtracting [latex]90x[/latex] from both sides,  [latex]180x - 90x - 300 = 90x - 90x - 120[/latex]
   [latex]90x - 300 = -120[/latex]
   Adding 300 to both sides,  [latex]90x - 300 + 300 = -120 + 300[/latex]
   [latex]90x = 180[/latex]
   Dividing both sides by 90,  [latex]\displaystyle{\frac{90x}{90} = \frac{180}{90}}[/latex]
   [latex]x = 2[/latex]
5. [latex]\displaystyle{\frac{x + 2}{3} = \frac{5 - 2x}{7}}[/latex]
   Cross-multiplying,  [latex]7(x + 2) = 3(5 - 2x)[/latex]
   Expanding both sides,[latex]7x + 14 = 15 - 6x[/latex]
   Adding [latex]6x[/latex] to both sides,  [latex]7x + 6x + 14 = 15 - 6x + 6x[/latex]
   [latex]13x + 14 = 15[/latex]
   Subtracting 14 from both sides, [latex]13x + 14 - 14 = 15 - 14[/latex]
   [latex]13x = 1[/latex]
   Dividing both sides by 13,  [latex]\displaystyle{\frac{13x}{13} = \frac{1}{13}}[/latex]
   [latex]\displaystyle{x = \frac{1}{13}}[/latex]

Steps to Solve Word Problems

Step 1:

Read the entire problem and ensure you understand the situation.

Step 2:

Identify the given information and the question to be answered.

Step 3:

Look for keywords. Some words indicate certain mathematical operations (see Table 7.1).

Step 4:

Choose a variable to represent the unknown(s) and state what that variable represents, including the unit of measure.

Note: For now, if there is more than one unknown, try to identify all the unknowns in terms of one variable, as all the questions in this chapter can be solved with only one variable.

Step 5:

Where necessary, draw a simple sketch to identify the information. This helps with envisioning the question more clearly.

Step 6:

Create an equation (or set of equations) to describe the relationship between the variables and the constants in the question.

Step 7:

Group like terms, isolate the variable and solve for the unknown(s).

Step 8:

State the solution to the given problem.

Solution

Let Harry’s age today be x years.

Therefore, in 5 years, Harry’s age will be:

[latex]x + 5 = 65[/latex]   Solving for [latex]x[/latex],

[latex]x = 65 - 5 = 60[/latex]

Therefore, Harry is 60 years old today.

Solution

Let the width be [latex]w[/latex] metres.

Therefore, the length is (w + 5) metres.

Perimeter = 2(length) + 2(width)

[latex]50 = 2(w + 5) + 2w[/latex]

[latex]50 = 2w + 10 + 2w[/latex]

[latex]2w + 10 + 2w = 50[/latex]

[latex]4w + 10 = 50[/latex]

[latex]4w = 50 - 10[/latex]

[latex]4w = 40[/latex]

[latex]\displaystyle{w = \frac{40}{4}}[/latex]

[latex]w = 10[/latex]

Therefore, the width of the garden is 10 metres and the length is (10 + 5) = 15 metres.

Solution

Let the cost of the Blu-ray player be [latex]\$x[/latex].

Therefore, the cost of the TV is [latex]\$(x + 190.00)[/latex].

The total cost is $688.00.

[latex]x + (x + 190.00) = 688.00[/latex]

[latex]x + x + 190.00 = 688.00[/latex]

[latex]2x + 190.00 = 688.00[/latex]

[latex]2x = 688.00 - 190.00[/latex]

[latex]2x = 498.00[/latex]

[latex]\displaystyle{x = \frac{498.00}{2}}[/latex]

[latex]x = \$249.00[/latex]

Therefore, the cost of the Blu-ray player is $249.00 and the cost of the TV is (249.00 + 190.00) = $439.00.

Solution

From the last column, we get the equation for the saline mix. The number of litres of saline in the 15% solution must be the same as the number of litres in the final 10% solution, as only water is being added, which does not contribute any additional saline to the solution. Therefore,

[latex]4.5 = 0.10 \times (30 + x)[/latex]

[latex]4.5 = 3 + 0.10x[/latex]

[latex]1.5 = 0.10x[/latex]

[latex]x = 15[/latex]

Therefore, 15 litres of water need to be added to the 15% saline solution to make the solution 10% saline.

7.3 Exercises

Answers to the odd-numbered problems are available at the end of the textbook.

For problems 1 to 8, simplify and evaluate the expressions.

1. The sum of a number and six is ten.
2. A number decreased by fifteen is five.
3. Six times a number is seventy-two.
4. The product of a number and four is twenty-eight.
5. A number divided by five is four.
6. A number divided by three is three.
7. Two-thirds of a number is twelve.
8. Two-fifths of a number is six.

For problems 9 to 30, solve the algebraic equations using the properties of equality, and express the answer as a fraction in its lowest terms or as a mixed number, wherever applicable.

9. [latex]x - 20 = 10[/latex]
10. [latex]x - 25 = 17[/latex]
11. [latex]22 = 40 - x[/latex]
12. [latex]54 = 23 - x[/latex]
13. [latex]21 + x = 4[/latex]
14. [latex]50 + x = 45[/latex]
15. [latex]16 + x = 22[/latex]
16. [latex]12 + x = 38[/latex]
17. [latex]11x + 4 = 17[/latex]
18. [latex]7x - 16 = 22[/latex]
19. [latex]\displaystyle{x - \frac{4}{5} = \frac{3}{5}}[/latex]
20. [latex]\displaystyle{x - \frac{1}{6} = 1}[/latex]
21. [latex]\displaystyle{\frac{10}{15} = x - \frac{4}{3}}[/latex]
22. [latex]\displaystyle{\frac{x}{7} + 15 = 24}[/latex]
23. [latex]\displaystyle{x + \frac{2}{5} = \frac{1}{4}}[/latex]
24. [latex]\displaystyle{2x - \frac{2}{3} = \frac{5}{6}}[/latex]
25. [latex]5x = 20[/latex]
26. [latex]4x = 24[/latex]
27. [latex]\displaystyle{\frac{2x}{3} + 1 = \frac{5x}{8} + 2}[/latex]
28. [latex]\displaystyle{\frac{x}{2} - \frac{1}{6} = \frac{1}{3} + \frac{3x}{5}}[/latex]
29. [latex]\displaystyle{\frac{7x}{8} - 4 = \frac{x}{4} + 6}[/latex]
30. [latex]\displaystyle{\frac{8x}{3} - 5 = \frac{x}{3} + 2}[/latex]

For problems 31 to 54, solve the algebraic equations using the properties of equality, and round the answer to 2 decimal places, wherever applicable.

31. [latex]10y - 0.09y = 17[/latex]
32. [latex]x + 0.13x = 70[/latex]
33. [latex]0.3x - 3.2 = 0.4 - 0.6x[/latex]
34. [latex]4 + 0.2x = 0.7x - 0.5[/latex]
35. [latex]0.4x - 1.38 = 0.3x - 1.2[/latex]
36. [latex]1.2 - 0.7x = 2.7 - 0.5x[/latex]
37. [latex]0.43x + 0.25 = 0.29x - 0.03[/latex]
38. [latex]0.6x - 1.2 = 0.9 - 1.5x[/latex]
39. [latex]8x + 7 - 3x = -6x - 15 + x[/latex]
40. [latex]x - 2 - 4x = -3x - 8 + 5x[/latex]
41. [latex]2(3x - 7) = 28 - 3(x + 1)[/latex]
42. [latex]4(2x - 5) = 32 - 4(x - 2)[/latex]
43. [latex](4 + 6)(2 + 4x) = 45 - 2.5(x + 3)[/latex]
44. [latex](5 + 0.5x)(1 + 3) = -1.2(2x + 4) + 25[/latex]
45. [latex]15 + 5(x - 10) = 3(x - 1)[/latex]
46. [latex]2(x - 3) + 3(x - 5) = 4[/latex]
47. [latex]4(y + 7) - 2(y - 4) = 3(y - 2)[/latex]
48. [latex]8(2y + 4) - 6(3y + 7) = 3y[/latex]
49. [latex]\displaystyle{\frac{x - 7}{2} + \frac{x + 2}{3} = 41}[/latex]
50. [latex]\displaystyle{\frac{7}{12}(2x + 1) + \frac{3}{4}(x + 1) = 3}[/latex]
51. [latex]\displaystyle{\frac{5}{y + 4} = \frac{3}{y - 2}}[/latex]
52. [latex]\displaystyle{\frac{3}{x + 1} = \frac{2}{x - 3}}[/latex]
53. [latex]\displaystyle{\frac{7}{5x - 3} = \frac{5}{4x}}[/latex]
54. [latex]\displaystyle{\frac{5}{y + 2} = \frac{3}{y}}[/latex]

For problems 55 to 76, solve the word problems using algebraic equations.

55. If three times a number plus twenty is seven times that number, what is the number?
56. Fifteen less than three times a number is twice that number. What is the number?
57. A 25-metre-long wire is cut into two pieces. One piece is 7 metres longer than the other. Find the length of each piece.
58. A 9-metre-long pipe is cut into two pieces. One piece is twice the length of the other piece. Find the length of each piece.
59. $500 is shared between Andy and Becky. Andy’s share is $150 less than Becky’s share. Calculate the amount of each of their shares.
60. $200 is shared between Bill and Ann. Ann’s share is $50 more than Bill’s share. Calculate the size of each of their shares.
61. Movie tickets that were sold to each child were $3 cheaper than those sold to each adult. If a family of two adults and two children paid $34 to watch a movie at the cinema, what was the price of each adult ticket and each child ticket?
62. Giri had twice the number of quarters (25 cents) in his bag than dimes (10 cents). If he had a total of 54 coins, how many of them were quarters? What was the total dollar value of these coins?
63. A square garden, with sides of length x metres, is widened by 4 metres and lengthened by 3 metres. Write the equation for the area (A) of the expanded garden. If each side was originally 10 metres in length, find the new area. (Hint: Area of a Rectangle = Length × Width)
64. A square garden, with sides of length x metres, has had its width reduced by 4 metres and its length reduced by 2 metres. Write the equation for the Area (A) of the smaller garden. If each side was originally 20 metres in length, find the new area.
65. Aran bought a shirt and a pair of pants for $34.75. The pair of pants cost $9.75 more than the shirt. Calculate the cost of the shirt.
66. Mythili bought a schoolbag and a toy for $30.45. The school bag cost $5.45 more than the toy. Calculate the cost of the school bag.
67. Sam is paid $720 a week. He worked 9 hours of overtime last week and he received $954. Calculate his overtime rate per hour.
68. Lisa is paid $840 a week. Her overtime rate is $28 per hour. Last week she received $1,036. How many hours of overtime did she work last week?
69. The sum of the three angles of any triangle is 180°. If [latex]3x[/latex], [latex]7x[/latex], and [latex]8x[/latex] are the measures of the three angles of a triangle, calculate the measure of each angle of the triangle.
70. The sum of the three angles of any triangle is 180°. If [latex]3x[/latex], [latex]4x[/latex], and [latex]5x[/latex] are the measures of the three angles of a triangle, calculate the measure of each angle of the triangle.
71. The perimeter of a triangle is the sum of the lengths of the three sides of the triangle. The perimeter of a triangle with sides [latex]x[/latex] cm, [latex](x + 10)[/latex] cm, and [latex]2x[/latex] cm is 70 cm. Calculate the length of each side of the triangle.
72. The perimeter of a triangle is the sum of the lengths of the three sides of the triangle. The perimeter of a triangle with sides [latex]x + 10[/latex], [latex]2x + 10[/latex], and [latex]3x[/latex] is 110 cm. Calculate the length of each side of the triangle.
73. After completing a weight-loss program, a patient weighs 160 lb. His dietician observes that the patient has lost 15% of his original weight. What was the patient’s starting weight?
74. A beaker in a chemistry lab contains 3 litres of water. While conducting an experiment, the chemistry professor removes three-fifths of the water from the beaker. He then adds three-fifths of the remaining volume to the beaker. How much water is left in the beaker at the end of the experiment?
75. A researcher wants to make 4 L of a 7% acid solution. She has a beaker of 15% acid solution in stock. How much of the 15% solution does she need to use and how much water must she add in order to prepare her desired solution?
76. A chemist wants to make a 10% acid solution. She has 5 L of 25% acid solution. How many litres of water should she add to the 25% solution in order to prepare her desired solution?