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.

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  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.

To solve the system of equations,  and , 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 is the easier equation to solve for one variable, we’ll start there. Solve for . 

add  to both sides.

From here can insert this equation into all ’s in our other equation and solve

    distribute

       combine like terms

               subtract 9 from both sides

               divide by 6

From here we reinsert this value into our equation.

Solution:

Adding the two equations together gives , so .  Substituting  into one of the original equations gives

The sum of  and  is

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