$\displaystyle 3y+6x=24$

$\displaystyle 2y+x=16$

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.

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

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

$\displaystyle \dpi{100} \small y=9$

$\displaystyle 4x+8y=12$

$\displaystyle 5x+2y=15$

Multiply the bottom equation by $\displaystyle -4$

Add the two equations together. This will allow you to cancel out $\displaystyle y$ from an equation. From here, you can proceed to solve for $\displaystyle x$.

$\displaystyle 5x+2y=15$

$\displaystyle -20x-8y=-60$

$\displaystyle -16x=-48$

$\displaystyle x=3$

From here, plug in the $\displaystyle x$ value into the equation of your choice to solve for $\displaystyle y$.

$\displaystyle 4(3)+8y=12$

$\displaystyle y=0$

We are essentially presented with a system of equations. To solve for y, we will need to solve the system. The easiest way to solve this particular system is by adding the equations together.

First, multiply the second equation by 2.

$\displaystyle 2(-x + 5y = 8)$

$\displaystyle -2x+10y=16$

Adding the two equations together will allow you to cancel the x values and solve for y.

          $\displaystyle 2x + 3y = 10$

$\displaystyle +\ -2x+10y=16$

$\displaystyle 0x+13y=26$

$\displaystyle 13y=26$

$\displaystyle y=2$

If y equals 2, then 4y will be equal to 8.

To solve this problem, we translate the given information into two equations and then solve both simultaneously. If we let F represent the number of tanks that hold four fish and S represent the number of tanks that hold six fish, the problem tells us that F+S=15. The problem also tells us that 4F (the total number of fish in the 4-fish tanks) plus 6S (the total number of fish in the six-fish tanks) equals 70 (the total number of fish in the aquarium).

Thus we have the following system of equations:

F+S=15

4F+6S=70

Multiplying the first equation by -4 and combing it with the second gives 2S = 10, as seen below:

[-4F-4S=-60 (the first equation times -4)]

2S = 10

Therefore, S, the number of tanks that hold 6 fish, is 5.

Let $\displaystyle c$ be the distance from home to church

Let $\displaystyle m$ be the distance from church to mosque

$\displaystyle c = 2m - 3$

$\displaystyle c + m = 54$

$\displaystyle 2m - 3 + m = 54$

$\displaystyle m = 19$

$\displaystyle c = 35$

To solve for y, first eliminate x by adding the two equations together such that the x’s factor out:

4x + 9y + 7 = 0

(–2)2x – (–2) 3y + (–2) 6 = (–2)0 (Multiply this equation by a factor of –2 so that 4x – 4x = 0)

Therefore, the two equations added together are:

(4x + 9y + 7 = 0) + (–4x –(–6)y +(–12) = 0) = (0 + 15y – 5 = 0)

y = 1/3

Let $\displaystyle x$ be the number of dimes Julie has and $\displaystyle y$ be the numbers of quarters she has. The number of dimes and the number of quarters add up to $\displaystyle 35$ coins. The value of all quarters and dimes is $\displaystyle \$7.25$. We can then write the following system of equations:

$\displaystyle x+y=35$

$\displaystyle 0.1x+0.25y=7.25$

To use substitution to solve the problem, begin by rearranging the first equation so that $\displaystyle x$ is by itself on one side of the equals sign:

$\displaystyle x=35-y$

Then, we can replace $\displaystyle x$ in the second equation with $\displaystyle 35-y$:

$\displaystyle 0.1(35-y)+0.25y=7.25$

Distribute the $\displaystyle 0.1$:

$\displaystyle 3.5-0.1y+0.25y=7.25$

Subtract $\displaystyle 3.5$ from each side of the equation:

$\displaystyle 0.15y=3.75$

Divide each side of the equation by $\displaystyle 0.15$:

$\displaystyle y=25$

Now, we can insert our value for $\displaystyle y$ into the first equation and solve for $\displaystyle x$:

$\displaystyle x=35-y=35-25=10$

Julie has $\displaystyle 25$ quarters and $\displaystyle 10$ dimes.