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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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 Addition/Subtraction Method for Solving Systems of Two Equations Algebraically manipulate both equations so that all the variables are on one side of the equal sign and in the same order. (Line the equations up, one on top of the other.) If needed, multiply one of the equations by a constant so that there is one variable in each equation that has the same coefficient. Subtract one equation from the other. Solve the resulting equation for the one variable. (Use the technique described in previous units.) Using the value found in the step 4, substitute it into either equation and solve for the remaining variable. Substitute the values for both variables into the equation not used in step 5 to be sure our solution is correct.

While there seems to be a lot of steps to this process, itreally is rather simple. Let’s work through one example tosee how the steps work. We have two equations below:

P = 2 + 2Q
P
= 10 – 6Q

To solve for the numeric value of each variable, we shouldfollow the steps outlined above.

1. Algebraically manipulate both equationsso that all the variables are on one side of the equal sign andin the same order. (Line the equations up, one on top of the other.)

 When we do this we get the equations shown at the right. Notice how the terms containing P are lined up, as are the terms containing Q. P – 2Q = 2 P + 6Q = 10

1. If needed, multiply one of the equationsby a constant so that there is one variable in each equation thathas the same coefficient.

The variable P in both equations has a coefficient ofone, so we move on to the next step.

1. Subtract one equation from the other. 1. Solve the resulting equation for the one variable. 1. Using the value found in the step4, substitute it into either equation and solve for the remainingvariable.

 In the previous step we found that Q = 1. We plug this value back into either of our initial equations and solve for P. P = 2 + 2Q P = 2 + 2(1) P = 4

1. Substitute the values for both variablesinto the equation not used in step 5 to be sure our solution iscorrect.

 Now we substitute the value of Q = 1 and P = 4 into the equation not used in step 5. P = 10 – 6Q 4 = 10 – 6(1) 4 = 4.

These values work in this equation so we have the correct answer.

Keep in mind that you can use either method for solving systemsof equations. Let’s revisit the example we used in the lastsection, but this time solve it using addition/subtraction.

Example
Solve the system of linear equations given below usingaddition/subtraction.

Suppose there is a piggybank that contains 57 coins, whichare only quarters and dimes. The total number of coins in thebank is 57, and the total value of these coins is \$9.45. Thisinformation can be represented by the following system of equations:

D + Q = 57
0.10D + 0.25Q = 9.45

Determine how many of the coins are quarters and how many aredimes.

The information in this example can be represented by the followingsystem of equations:

D + Q = 57
0.10D + 0.25Q = 9.45
 Determine how many of the coins are quarters and how many are dimes. D = 32 (number of dimes) Q = 25 (number of quarters)

W
e were given the two equations:

D + Q = 57     (1)
0.10D + 0.25Q = 9.45     (2)

1. Algebraically manipulate both equationsso that all the variables are on one side of the equal sign andin the same order.

 When we do this we get the equations shown at the right. Notice how the terms containing D are lined up, as are the terms containing Q. D + Q = 57 0.10D + 0.25Q = 9.45

2. If needed, multiply one of the equationsby a constant so that there is one variable in each equation thathas the same coefficient.

Because neither of the variables has the same coefficient inboth equations, we will need to multiply one equation by a numberthat will result in the variable in one equation having the samecoefficient as in the other. In this example we can see the coefficientof Q in equation (1) is 4 times larger than the coefficient ofQ in equation (2).

 If we multiply the second equation by 4, then the variable Q will have the same coefficient in both equations. This means that both sides of equation (2) must be multiplied by 4. 0.10D + 0.25Q = 9.45 4(0.10D + 0.25Q = 9.45) 4(0.10)D + 4(0.25)Q = 4(9.45) 0.40D + 1Q = 37.8

3. Subtract one equation from the other. Now we subtract one equation from the other.

D + Q = 57
- (0.40D + Q = 37.8)

0.60D = 19.2

4. Solve the resulting equation for theone variable.

 We now solve the equation from step 3 to find a value for the D variable. 5. Using the value found in the step4, substitute it into either equation and solve for the remainingvariable.

 Now we substitute the value of D = 32 into one of the equations to solve for Q. D + Q = 57 32 + Q – 32= 57 – 32 Q = 25

6. Substitute the values for both variablesinto the equation not used in step 5 to be sure our solution iscorrect.

 Now we substitute the value of D = 32 and Q = 25 into the equation not used in step 5. 0.10D + 0.25Q = 9.45 .10(32) + .25(25) = 9.45 3.20 + 6.25 = 9.45 9.45 = 9.45

So we find that Q = 25 and D = 32. If you lookback at the last section you will find this is the same resultwe found then.

If you wish to try a practice problem now, return to the introductionand select the practice or additional practice button.