Linear equations  (Page 5/6)

We let $x=\text{the number of items manufactured}$ , and let $y=\text{the cost}$ .

Solving this problem is equivalent to finding an equation of a line that passes through the points (25, 750) and (50, 1000).

$m=\frac{\text{1000}-\text{750}}{\text{50}-\text{25}}=\text{10}$

Therefore, the partial equation is $y=\text{10}x+b$

By substituting one of the points in the equation, we get $b=\text{500}$

Therefore, the cost equation is $y=\text{10}x+\text{500}$

Now to find the cost of 100 items, we substitute $x=\text{100}$ in the equation $y=\text{10}x+\text{500}$

So the $\text{cost}=y=\text{10}(\text{100})+\text{500}=\text{1500}$

Let us look at what is given.

Again, solving this problem is equivalent to finding an equation of a line that passes through the points (0, 32) and (100, 212).

Since we are finding a linear relationship, we are looking for an equation $y=\text{mx}+b$ , or in this case $F=\text{mC}+b$ , where $x$ or $C$ represent the temperature in Celsius, and $y$ or $F$ the temperature in Fahrenheit.

The equation is $F=\frac{9}{5}C+b$

Substituting the point (0, 32), we get

Now to convert 30 degrees Celsius into Fahrenheit, we substitute $C=\text{30}$ in the equation

The problem can be made easier by using 1970 as the base year, that is, we choose the year 1970 as the year zero. This will mean that the year 1986 will correspond to year 16, and the year 2010 as the year 40.

Now we look at the information we have.

Solving this problem is equivalent to finding an equation of a line that passes through the points (0, 18) and (16, 26).

The equation is $y=\frac{1}{2}x+b$

Substituting the point (0, 18), we get

Now to find the population in the year 2010, we let $x=\text{40}$ in the equation

So the population of Canada in the year 2010 will be 38 million.

We graph both lines on the same axes, as shown below, and read the solution (2, 5).

Finding an intersection of two lines graphically is not always easy or practical; therefore, we will now learn to solve these problems algebraically.

At the point where two lines intersect, the x and y values for both lines are the same. So in order to find the intersection, we either let the x-values or the y-values equal.

If we were to solve the above example algebraically, it will be easier to let the y-values equal. Since $y=\mathrm{3x}-1$ for the first line, and $y=-x+7$ for the second line, by letting the y-values equal, we get

By substituting $x=2$ in any of the two equations, we obtain $y=5$ .

Hence, the solution (2, 5).

One common algebraic method used in solving systems of equations is called the elimination method . The object of this method is to eliminate one of the two variables by adding the left and right sides of the equations together. Once one variable is eliminated, we get an equation that has only one variable for which it can be solved. Finally, by substituting the value of the variable that has been found in one of the original equations, we get the value of the other variable. The method is demonstrated in the example below.