Jimmy starts off with $60, and spends $3.50 everyday.  

This means that he will have $56.50 after day 1, $53 after day 2, and so forth.  

Only one equation satisfies this scenario.  The rest are irrelevant.

$\displaystyle T= 60-3.50D$

We add the two systems of equations:

For the Left Hand Side:

$\displaystyle (8x+3y)+(2x-3y)=10x$

For the Right Hand Side:

$\displaystyle 2+18=20$

So our resulting equation is:

$\displaystyle 10x=20$

Divide both sides by 10:

For the Left Hand Side:

$\displaystyle \frac{10x}{10}=x$

For the Right Hand Side:

$\displaystyle \frac{20}{10}=2$

Our result is:

$\displaystyle x=2$

Reduce the second system by dividing by 3.

Second Equation:

$\displaystyle 12x+6y=24$     We this by 3.

$\displaystyle \frac{12x}{3}+\frac{6y}{3}=\frac{24}{3}$

$\displaystyle 4x+2y=8$

Then we subtract the first equation from our new equation.

First Equation:

$\displaystyle 5x+2y=9$

First Equation - Second Equation:

Left Hand Side:

$\displaystyle (5x+2y)-(4x+2y)=x$

Right Hand Side:

$\displaystyle 9-8=1$

Our result is:

$\displaystyle x=1$

We first add both systems of equations.

Left Hand Side:

$\displaystyle (2x-y)+(x+y)=3x$

Right Hand Side:

$\displaystyle 2+4=6$

Our resulting equation is:

$\displaystyle 3x=6$

We divide both sides by 3.

Left Hand Side:

$\displaystyle \frac{3x}{3}=x$

Right Hand Side:

$\displaystyle \frac{6}{3}=2$

Our resulting equation is:

$\displaystyle x=2$

We first multiply the second equation by 4.

So our resulting equation is:

$\displaystyle x\cdot4+4y\cdot4=17\cdot4$

$\displaystyle 4x+16y=68$

Then we subtract the first equation from the second new equation.

Left Hand Side:

$\displaystyle (4x+y)-(4x+16y)=-15y$

Right Hand Side:

$\displaystyle 6-68=-60$

Resulting Equation:

$\displaystyle -15y=-60$

We divide both sides by -15

Left Hand Side:

$\displaystyle \frac{-15y}{-15}=y$

Right Hand Side:

$\displaystyle \frac{-60}{-15}=4$

Our result is:

$\displaystyle y=4$

We can express Dr. Jones's rate in a linear equation:

$\displaystyle \text{Cost}=$40(\text{number of 10 minute intervals})+50$

Since Mrs. Smith's appointment lasted 30 minutes, we have 3 10-minute intervals. Then, we can plug in that number into our above equation to find out how much the appointment cost.

$\displaystyle \text{Cost}=40(3)+50=120+50=170$

The total number of uniforms needed equals the number of students divided by the number of uniforms per box.

First, add 6 to both sides so that the term with "x" is on its own.

$\displaystyle 2x=20$

Now, divide both sides by 2.

$\displaystyle x=10$

Start by isolating the term with $\displaystyle b$ to one side. Add 10 on both sides.

$\displaystyle 7b=14$

Divide both sides by 7.

$\displaystyle b=2$

First start by distributing the 7.

$\displaystyle 7(t-2)=7t-14$

$\displaystyle 7t-14=21$

Now, add both sides by 14.

$\displaystyle 7t=35$

Finally, divide both sides by 7.

$\displaystyle t=5$