Linear equations  (Page 6/6)

We add the left and right sides of the two equations.

Now we substitute $x=2$ in any of the two equations and solve for $y$ .

Therefore, the solution is (2, 3).

If we add the two equations, none of the variables are eliminated. But the variable $x$ can be eliminated by multiplying the first equation by –2, and leaving the second equation unchanged.

Substituting $y=2$ in $x+\mathrm{2y}=3$ , we get

Therefore, the solution is (–1, 2).

This time, we multiply the first equation by – 4 and the second by 3 before adding. (The choice of numbers is not unique.)

By substituting $y=-2$ in any one of the equations, we get $x=-1$ . Hence the solution (–1, –2).

1. We substitute $x=\text{10}$ in the supply equation, $y=1\text{.}\mathrm{5x}+\text{10}$ , and the answer is $y=\text{25}$ .
2. We substitute $x=\text{10}$ in the demand equation, $y=-2\text{.}\mathrm{5x}+\text{34}$ , and the answer is $y=9$ .

3. By letting the supply equal the demand, we get

   $1\text{.}\mathrm{5x}+\text{10}=-2\text{.}\mathrm{5x}+\text{34}$
   $\mathrm{4x}=\text{24}$
   $x=6$

4. We substitute $x=6$ in either the supply or the demand equation and we get $y=\text{19}$ .

   The graph below shows the intersection of the supply and the demand functions and their point of intersection, (6, 19).

   

1. We substitute $x=4$ in the revenue equation $R=\mathrm{5x}$ , and the answer is $R=\text{20}$ .
2. We substitute $x=4$ in the cost equation $C=\mathrm{3x}+\text{12}$ , and the answer is $C=\text{24}$ .

3. By letting the revenue equal the cost, we get

   $5x=3x+12$
   $x=6$

4. We substitute $x=6$ in either the revenue or the cost equation, and we get $R=C=30$ .

   The graph below shows the intersection of the revenue and the cost functions and their point of intersection, (6, 30).