Plugging in –3 for n gives a system that, when added vertically, gives 0 = 24, which is untrue.

Add the two equations to get 3x = 9, so x = 3.  Substitute the value of x into one of the equations to find the value of y; therefore x = 3 and y = -2, so their sum is 1.

Pure water is considered 0% whereas pure solution is 100%.

The general equations is V_water x P_water + V_soultion x P_solution = V_final x P_final where

V means volume and P means percent.

x(0) + 1(1.00) = (x + 1)(0.60) and solve for x = volume of pure water.

Define variables as x = Billy's age and x + 4 = Joey's age

The sum of their ages is x + (x + 4) = 24

Solving for x, we get that Billy is 10 years old and Joey is 14 years old.

Solution A:

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

     \(\displaystyle 3a+2b=16\)

\(\displaystyle -(3a-2b=4)\)  (don't forget to distribute the minus sign throughout!)

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\(\displaystyle 4b=12\)

\(\displaystyle b=3\)

Solution B:

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

\(\displaystyle 3a+2b=16\)

\(\displaystyle 3a=16-2b\)

\(\displaystyle a=\frac{16-2b}{3}\)

\(\displaystyle 3(\frac{16-2b}{3})-2b=4\)

\(\displaystyle (16-2b)-2b=4\)

\(\displaystyle -4b=-12\)

\(\displaystyle b=3\)

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.

To solve the system of equations, \(\displaystyle 3x+3y=15\) and \(\displaystyle y-x=3\), we must begin by inserting one equation into another. First notice that there are two variables and that we have two equations, therefore, we have enough equations to solve for both variables. As an aside, for each variable, you need that many equations to solve for all variables.

We will insert the equations by substitution, which tends to be a more useful form of equation integration. As the \(\displaystyle y-x=3\) is the easier equation to solve for one variable, we’ll start there. Solve for \(\displaystyle y\). 

\(\displaystyle y-x=3\) add \(\displaystyle x\) to both sides.

\(\displaystyle y=3+x\)

From here can insert this equation into all \(\displaystyle y\)’s in our other equation and solve

\(\displaystyle 3x+3(3+x)=15)\)    distribute

\(\displaystyle 3x+9+3x=15\)       combine like terms

\(\displaystyle 6x+9=15\)               subtract 9 from both sides

\(\displaystyle 6x=15-9\)               divide by 6

\(\displaystyle x=\frac{6}{6}\)

\(\displaystyle x=1\)

From here we reinsert this value into our \(\displaystyle y=3+x\) equation.

\(\displaystyle y=3+1\)

\(\displaystyle y=4\)

Solution: \(\displaystyle \dpi{100} x=1,y=4\)

Adding the two equations together gives \(\displaystyle 5x = 15\), so \(\displaystyle x = 3\).  Substituting \(\displaystyle x\) into one of the original equations gives \(\displaystyle y = 4\)

The sum of \(\displaystyle x\) and \(\displaystyle y\) is \(\displaystyle 7\)

Factor out this binomial. –3 and 2 are the only possible x values. 2 is the answer.

Use substitution and solve for one variable, then back substitute and solve for the other variable, or use elimination. 

There are two ways to solve this problem.

Option 1: The Substitution Method

Step 1: Set up the second equation so that \(\displaystyle y\) is by itself

\(\displaystyle 4x+y=22\rightarrow y=22-4x\)

Step 2: Substitute for \(\displaystyle y\) in the first equation, and solve for \(\displaystyle x\)

\(\displaystyle 3x+2(22-4x)=29\)

\(\displaystyle 3x+44-8x=29\)

\(\displaystyle -5x=-15\)

\(\displaystyle x=3\)

Step 3: Plug \(\displaystyle 3\) into the second equation and solve for \(\displaystyle y\)

\(\displaystyle 4(3)+y=22\)

\(\displaystyle 12+y=22\)

\(\displaystyle y=10\)

Option 2: The Elimination Method

Step 1: Set up the equations so that the variables are on the same side

\(\displaystyle 3x+2y=29\)

\(\displaystyle 4x+y=22\)

Step 2: Multiple the second equation by 2

\(\displaystyle 3x+2y=29\)

\(\displaystyle 8x+2y=44\)

Step 3: Subtract the second equation from the first (thereby canceling out the \(\displaystyle y\)s) and solve for x

\(\displaystyle 3x+2y=29\)

\(\displaystyle 8x+2y=44\)

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\(\displaystyle -5x=-15\)

\(\displaystyle x=3\)

Step 4: Substitute \(\displaystyle 3\) into one of the equations and solve for \(\displaystyle y\)

\(\displaystyle 3(3)+2y=29\)

\(\displaystyle 9+2y=29\)

\(\displaystyle 2y = 20\)

\(\displaystyle y=10\)