The break-even point is where the costs equal the revenue.

Costs:  \(\displaystyle C(b)=600+35b\)

Revenue:  \(\displaystyle R(b)=75b\)

So Costs = Revenue or \(\displaystyle 600+35b=75b\)

Solving for \(\displaystyle b\) shows the break-even point is achieved when 15 pairs of boots are sold in a month.

Profits = Revenue - Costs

Revenue:  \(\displaystyle R(b)=75b\)

Costs:  \(\displaystyle C(b)=600+35b\)

So

\(\displaystyle P(b)=R(b)-C(b)=75b-(600+35b)=40b-600\)

Solving:

  \(\displaystyle 200=40b-600\)

for \(\displaystyle b\) means that 20 pairs of boots must be sold to make $200 profit.

This problem can be solved by using substitution.  Write the first equation  in terms of \(\displaystyle y\) and substitute it into the second equation.

So \(\displaystyle y = \frac{19}{3} - \frac{2}{3}x\) and thus \(\displaystyle 3x + 2(\frac{19 }{3}-\frac{2}{3}x) = 16\) and solving for \(\displaystyle x = 2\) and then \(\displaystyle y = 5\).

So the sum of \(\displaystyle x\) and \(\displaystyle y\) is 7.

We have 3 unkown variables and only 2 equations. Instead of trying to solve for \(\displaystyle x\)  or \(\displaystyle y\), notice that we can substitute 2 in for the entire expression \(\displaystyle (2x+y)\):

\(\displaystyle 4x+3z+8y = 4x + 8y + 3z = 4(x+2y) + 3z = 26\)

Substitution gives:

\(\displaystyle 4(2)+3z =26\)

Subtract 8 from both sides: 

\(\displaystyle 3z = 18\)

Divide by 3:

\(\displaystyle z=6\)

We need 2 equations, because we have 2 unkown variables. Let \(\displaystyle e\) = the entrance fee, and \(\displaystyle r\) = the fee per ride. One ride costs \(\displaystyle r\) dollars. We know that Lisa spent 120 dollars in total. Since Lisa went on 6 rides, she spent \(\displaystyle 6r\) dollars on rides. Her only other expense was the entrance fee, \(\displaystyle e\): 

\(\displaystyle 120 =6r+e\)

Apply similar logic to Tom:

\(\displaystyle 95=4r+e\)

Subtracting the second equation from the first equation results in:

\(\displaystyle 25 = 2r\)

Divide both sides by 2:

\(\displaystyle r = 12.5\)

So every ride costs 12.5 dollars. Plugging 12.5 back into one of the original equations allows us to solve for the entrance fee:

\(\displaystyle 95 = 4(12.5) +e\)

\(\displaystyle 95 = 50 +e\)

Subtract 50 from both sides:

\(\displaystyle e = 45\)

\(\displaystyle 4x+6y=2\)

\(\displaystyle y=2x\)

For this system, it will be easiest to solve by substitution. The \(\displaystyle y\) variable is already isolated in the second equation. We can replace \(\displaystyle y\) in the first equation with \(\displaystyle 2x\), since these two values are equal.

\(\displaystyle 4x+6y=2\)

\(\displaystyle 4x+6(2x)=2\)

Now we can solve for \(\displaystyle x\).

\(\displaystyle 4x+12x=2\)

\(\displaystyle 16x=2\)

\(\displaystyle x=\frac{2}{16}=\frac{1}{8}\)

Now that we know the value of \(\displaystyle x\), we can solve for \(\displaystyle y\) by using our original second equation.

\(\displaystyle y=2x\ \text{and}\ x=\frac{1}{8}\)

\(\displaystyle y=2(\frac{1}{8})\)

\(\displaystyle y=\frac{2}{8}=\frac{1}{4}\)

The final answer will be the ordered pair \(\displaystyle (\frac{1}{8}, \frac{1}{4})\).

\(\displaystyle 3y+6x=24\)

\(\displaystyle 2y+x=16\)

Solve the system of equations using substitution.

First, isolate one of the variables. Since we are solving for \(\displaystyle y\), we are going to isolate \(\displaystyle x\) in the second equation. 

\(\displaystyle 2y+x=16\)

\(\displaystyle x=-2y+16\)

Replace \(\displaystyle x\) with \(\displaystyle -2y+16\) in our first equation.

\(\displaystyle 3y+6x=24\)

\(\displaystyle 3y+6(-2y+16)=24\)

Now we can solve to isolate \(\displaystyle y\).

\(\displaystyle 3y+(-12y+96)=24\)

\(\displaystyle -9y+96=24\)

\(\displaystyle -9y=-72\)

\(\displaystyle y=8\)

After 20 minutes the first train would have traveled 20 miles. Let x be the amount of time elapsed. When 20 + 60x = 100x you will have the time in hours. 20 = 40x, x = 0.5 hrs. 0.5 hrs = 30 minutes. 

Solve by method of elimination. Multiply the first equation by 8 to eliminate the variable, m. Our first equation will then become \(\displaystyle \dpi{100} \small -4m+8n=24\).

By adding this new equation

\(\displaystyle \dpi{100} \small -4m+8n=24\)

with our second equation

\(\displaystyle \dpi{100} \small 4m+2n=16\)

We will see that our m cancel out. We can now solve for n. 

\(\displaystyle \dpi{100} \small 10n=40\)

\(\displaystyle \dpi{100} \small n=4\)

Now we have to plug in this value of n into any of our equations to find the value of m. 

Let's use the second equation.

\(\displaystyle \dpi{100} \small 4m+2\left (4 \right )=16\)

\(\displaystyle \dpi{100} \small 4m+8=16\)

\(\displaystyle \dpi{100} \small 4m=8\)

\(\displaystyle \dpi{100} \small m=2\)

\(\displaystyle \dpi{100} \small m-n=-2\)

Substitution needs to be used in order to solve this system of equations. From the second equation we know that \(\displaystyle \dpi{100} \small y=5+x\),

Substitute that into the first equation and solve.

You get \(\displaystyle \dpi{100} \small 10x - (5+x)=31\)

\(\displaystyle \dpi{100} \small 10x-5-x =31\)

\(\displaystyle \dpi{100} \small 9x =36\)

\(\displaystyle \dpi{100} \small x=4\)

From there solve for y using the second equation.

\(\displaystyle \dpi{100} \small y-x=5\)

\(\displaystyle \dpi{100} \small y-4=5\)

\(\displaystyle \dpi{100} \small y=9\)