Use elimination to solve for one variable.

y = 2

Back substitute for the other variable. 

x = 2

Substitute the second equation into the first or use elimination. 

Use substitution and plug in to solve for one equation. Then use back substitution to solve for the other variable.

Tom's allowance () is three times as much as Joe's ().

The sum of their allowance is $56.

Restate the second equation by substituting  for  to get . Solve for .

Add the two equations to get 

or 

and then substitute  into one of the original equations to get

Solving for  we get . 

And we then substitute  into one of the original equations to get . 

So the sum of  and  is .

The general equation to use is:

where  = coin value and  = number of coins.

Let  = number of quarters,  = number of dimes and  = number of nickels.

So the equation to solve becomes

Solving shows that Jacob had 10 nickels, 5 dimes and 5 quarters for a total of $2.25.

In 2010 the family sold number of apples.  

This increased by 100 in 2011: .  

In 2012 the number of apples tripled: 

In 2013 the number of apples increased by 200: 

If , we must solve for.

so in 2011 the number of apples sold was 500, or  apples.  

First we must solve for , then substitute into the other equation. Since we want in terms of , solve for  in the equation and substitute our value of  (in terms of ) into the equation, then simplify:

Now that we have , let's plug that into the equation.

Already we can see that this problem is a mess because it is an expression with two denominators. Remember that dividing by a number is equal to multiplying by that number's inverse. Thus, dividing by  is the same as multiplying by . So let's make an equivalent expression look like this:

This is much better as we can multiply straight across to get:

Now we can solve for .

One method of solving a system of equations requires multiplying one equation by a factor that will allow for the removal of one variable. In this system, we can multiply (x+y=5) by -2. When the -2 is distributed across the entire equation, the equation becomes (-2x-2y=-10). We then add the two equations: (-2x-2y=-10) + (2x+3y=20). When we do this, the x variable cancels out, leaving us with y=10. We then subsitute 10 for y in either of the original equations: (x+10=5) or (2x+ 30=20). Either way, we end up with x=-5. If you got an answer of 5, you may have made a computation error. If you got 10, you may have forgotten to substitute the y-value into one of the original equations to solve for x.

We are given st + 12= 3s + sv and t-v=7. Since we have a numerical value for t-v+, it would be ideal to isolate that in the first equation. We can do this be rearranging the first equation. Starting with st + 12= 3s + sv, we can subtract the 12 from the left giving us a -12 on the right. We then subtract the sv from the right and put it on the left. This gives us st-sv=3s-12. We then factor out the s on the left side of the equation, giving us s(t-v)=3s-12. We then plug-in 7 for t-v, giving us 7s=3s-12. We then solve for s, giving us -3.