If a number of equations involve the same variables, they are called a system of equations.

Example 2.5.1

An agent is to purchase two products, G and H, and send them to the company’s warehouse. He has a budget of $70,000 and truck space of 9,000 cubic meters. The product costs and sizes are as follows:

The question to be answered here is: how much of each product should be purchased if both the budget and space are to be used up?

Limitations such as those for budget and space are called constraints. Each constraint gives an equation. To find the equation from the data given, you should consider the totals for each constraint  separately and find the contribution of each  variable. In this case; the variables are the amounts of each product to be bought. Let,

•  g = the amount (number of cases) of G
•  h = the amount (number of cases) of H

Then the budget constraint equation can be obtained by considering:

[latex]\begin{align*} \text{Total Spent} &= \text{total spent on G} + \text{total spent on H}\\ &= \text{cost per case of G}\times\text{(amount of G) + cost per case of H}\times \text{(amount of H)}\\ &=$10\times g+$2\times h\\ $70,000 &=$10g +$20h\\ \end{align*}[/latex]

Similarly,

[latex]\begin{align*} \text{Total Space} &= \text{space used by G + total space used by H}\\ &= \text{space used by G} \times \text{(amount of G)+ space used by} \times \text{(amount of H)} \end{align*}[/latex]

So,

[latex]9,000 m^3=3g m^3+2h m^3[/latex]

Thus you have the system of equations

[latex]$70,000 = $10g + $20 h\text{ and } 9,000 = 3g + 2h[/latex]

for which the graphs are given below.

Notice that the data for the problem was chiefly in terms of rates, the costs per case and the volume per case. These were used to get equations for totals, using the idea of,

[latex]Amount = rate \times base[/latex]

for each equation. Notice, also, the use of the units of measurement to help keep track of the parts of the equations.

The values required by the problem  are those of the point at which both equations are satisfied – the point on the graph at which the  lines cross. This point is called the solution of the equations. It can be estimated from the graph and also calculated from the equations.

Method of Elimination

To solve the equations, using the method of elimination write each one in the form for which h is given as a function of g.

For the budget: h =  3,500 − 0.5g
For the space: h = 4,500 − 1.5g

Next, note that at the solution, the values of h must be the same, so:

[latex]3,500 - 0.5g = 4,500 - 1.5g[/latex]

By adding [latex]1.5g[/latex] to each side, you get:

[latex]3,500 - 0.5g + 1.5g = 4,500 - 1.5g + 1.5g[/latex]
[latex]3,500 + 1g= 4,500[/latex]

Subtracting $3,500 from each side gives:

[latex]g = 4,500 - 3,500[/latex]
[latex]g = 1,000 \text{ cases}[/latex]

Substituting the result in one of the original equations (space):

[latex]9,000 = 3\times 1,000 + 2h[/latex]

Whence

[latex]h = 6,000 + 2 = 3,000 \text{ cases}[/latex]

Thus, both the budget and space allowances will be used up if 1,000 cases of G and 3,000 cases of H are bought.

Method of Substitution

The method of substitution gives you another way of solving such equations. It consists of the following:

Once the first equation has been solved for h, the result is

[latex]h = 3,500 - 0.5g[/latex]

This result would be substituted in the second equation, which would produce an equation containing only the variable g.

Thus,

[latex]\begin{align*} 9,000 &= 3g + 2\times (3,500 - 0.5g)\\ 9,000 &= 3g + 7,000 - g \end{align*}[/latex]

and

[latex]\begin{align*} 9,000 - 7,000 &= 2g\\ 2,000 &= 2g\\ g &= 1,000 \end{align*}[/latex]

Substituting in the other equation:

[latex]\begin{align*} 70,000 &= 10 \times 1,000 + 20h\\ 70,000 -10,000 &= 20h\\ 60,000 &= 20h\\ h &= 3,000 \text{ cases} \end{align*}[/latex]

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