One can describe the ages of Jacob, Sarah, and Caroline with the letters J, S, and C, respectively. From the information in the problem, J = S + 3, and C = 2S. Since C = 28, S = 28/2 = 14, and J = 14 + 3 = 17. Jacob is 17 years old.

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.

One can describe the ages of Jacob, Sarah, and Caroline with the letters J, S, and C, respectively. From the information in the problem, J = S + 3, and C = 2S. Since C = 28, S = 28/2 = 14, and J = 14 + 3 = 17. Jacob is 17 years old.

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.

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

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 .

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

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 .

Solution A:

Notice that the two equations have very similar terms. If the two expressions are subtracted from each other, the variable cancels out:

(don't forget to distribute the minus sign throughout!)

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Solution B:

Using one of the equations, solve for a in terms of b:

Note: Solution A is the much faster way to solve this problem. Whenever you are asked to solve a problem with two equations and two variables (or more!), see if you can add them together or subtract them from each other to make the other variables cancel out.

Solution A:

Notice that the two equations have very similar terms. If the two expressions are subtracted from each other, the variable cancels out:

(don't forget to distribute the minus sign throughout!)

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Solution B:

Using one of the equations, solve for a in terms of b:

Note: Solution A is the much faster way to solve this problem. Whenever you are asked to solve a problem with two equations and two variables (or more!), see if you can add them together or subtract them from each other to make the other variables cancel out.