# Solving Linear Equations In Algebra

One of the goals in mathematics is solving equations. The most fundamental of these is linear equations. In this chapter, I discuss how to solve them.

**Introduction**

An equation is a statement that says two mathematical expressions are equal. Equations can have one or multiple variables. Solving an equation means finding all values for the variable that make the equation a true statement. The simplest type is the linear equation. A linear equation with one variable is an equation that can be written in the form: \( ax + b = 0 \).

**Example 1**

Solve the equation \( 3(x-4) = 2x - 1 \).

\( 3x - 12 = 2x - 1 \)

\( x = 11 \)

**Example 2**

Solve \( 3(2x - 5) = 10 - (x + 5) \).

\( 6x -15 = 10 -x -5 \)

\( 6x - 15 = 5 - x \)

\( 7x = 20 \)

\( x = \frac{20}{7} \)

When fractions or decimals are in an equation, you multiply each side of the equation by the least common denominator of all fractions in the equation.

**Example 3**

Solve: \( \frac{T-2}{4} - \frac{T}{3} = 5 - \frac{1}{12}(3 - T) \)

\( 3(T-2) - 40T = 60 - (3 - T) \)

\( 3T - 6 - 4T = 60 - 3 + T \)

\[(-2T = 63 \)

\( T = \frac{-63}{2} \)

**Example 4**

Solve: \( .03(z - 3) - .5(2z + 1) = .23 \)

To eliminate decimals in this case, multiple them by 100.

\( 3(z-3) - 50(2z+1) = 23 \)

\( 3z - 9 - 100z - 50 = 23 \)

\( -97z = 82 \)

\( z = \frac{-82}{97} \)

The equation \(f(x)\) = \(g(x)\) results whenever the formulas for two functions (f) and (g) are set equal to each other. A solution to this equation corresponds to the x-coordinate of a point where the graph of (f) and (g) intersect.

**Example 5**

Solve: \( 2x - 1 = \frac{1x}{2} + 2 \)

\( 2x = \frac{1}{2} x + 3 \)

\( \frac{3}{2}x = 3 \)

\( x = 2 \)

Linear equations and functions can be solved symbolically, graphically, and numerically. Symbolic solutions are always exact. Graphical and numerical solutions are approximated to some degree. The intermediate value property states that if two points are connected, then (f) assumes every value between the (y) points at least once.

**Example 6**

A survey found that 76% of bicycle riders do not wear helmets. Find a symbolic representation for a function that computes the number of people who do not wear helmets. Also, there are around 38.7 million riders who do not wear helmets. Write a linear equation that gives the total number of riders.

A linear function that computes 76% of (f) is: \( f(x) = .76x \).

We must find the x-value for which \( f(x) = 38.7 \). The equation is \( .76x = 38.7 \) which evaluates to \( x = \frac{38.7}{.76} \) and is equal to 50.9 million riders.

**Solving Problems**

Read the problem and make sure you understand it. Assign a variable to what you are being asked to find. Write an equation that relates the quantities described in the problem. Solve the equation and determine the solution. Check your solution and make sure it seems plausible.

**Example 7**

A large pump can empty a tank of gasoline in 5 hours and a smaller pump can empty the same tank in 9 hours. If both pumps are used to empty the tank, how long will it take?

We are looking for time, so let time equal (T). In one hour the large pump will empty \(\frac{1}{5}\) of the tank and the smaller pump will empty \(\frac{1}{9}\) of the tank. The fraction of the tank they will empty together in 1 hour is \( \frac{1}{5} + \frac{1}{9} \). So, in (T) hours, the fraction of the tank that the two pumps can empty is \(\frac{T}{5} + \frac{T}{9} \). Since the tank is empty when this function reaches 1, we can use \(\frac{T}{5} + \frac{T}{9} = 1 \).

\( \frac{T}{5} + \frac{T}{9} = 1 \)

\( \frac{45T}{5} + \frac{45T}{9} = 45 \)

\( 9T + 5T = 45 \)

\( 14T = 45 \)

\( T = \frac{45}{14} \)

\( T = 3.21 hours \)

**Example 8**

In one hour an athlete travels 10.1 miles by running 8 mph and then at 11 mph. How long did the athlete run at each speed?

We are asked to find the time spent running at each rate. If we let (x) represent the time spent running at 8 mph, then 1-x represents the time running at 11 mph because the total running time was 1 hour.

Distance (d) equals rate times time (T) or \( d=rt\). In this example, we have two rates and two times. The total distance must sum to 10.1 miles.

\( d = 10.1 = 8x + 11(1-x) \)

\( 10.1 = 8x + 11 -11x \)

\( 10.1 = 11-3x \)

\( 3x = .9 \)

\( x = .3 \)

So the athlete runs .3 of an hour at 8 mph and .7 of an hour at 11 mph.

**Example 9**

A person 6 feet tall stands 17 feet from the base of a streetlight. If the person’s shadow is 8 feet, estimate the height of the streetlight.

We are being asked to find the height of a streetlight, this will be (x).

You can use similar proportions here to solve this quickly.

\( \frac{x}{6} = \frac{25}{8} \)

\( x = \frac{(6)(25)}{8} \)

\(x = 18.75 ft. \)

The streetlight is 18.75 feet tall.

**Example 10**

Pure water is being added to a 30% solution of 153 ml of hydrochloric acid. How much water should be added to reduce it to a 13% mixture?

We are asked to find the amount of water that should be added to 153ml of 30% acid to make it a 13% solution. Let this amount of water equal x. Then x + 153 equals the final volume of the 13% solution.

Since only water is added, the total amount of acid in the solution after adding the water must equal the amount of acid before the water is added. The volume of pure acid after the water is added equals 13% of x + 153ml, and the volume of pure acid before the water is added equals 30% of 153ml. Our equation becomes:

\( .13(x + 153) = .30(153) \)

\( x + 153 = \frac{.30(153)}{.13} \)

\( x = \frac{.30(153) -153}{.13} \)

\( x = 200.08 ml \)

**Conclusion**

In this chapter I have talked about the fundamental linear equation. It is the easiest to understand and algebra essentials. You can see how many quantities are modeled and then estimated using basic algebra.