Systems of Equations
Sometimes an algebra problem includes more than one equation. Solving the homework problem involves solving the equations together. A problem of this type is called a system of equations. Systems of equations are found through out math and science and have real world applications. A system of linear equations is probably the easiest to solve although the same methods can be applied to a nonlinear system of equations. The methods most frequently used are graphing, substitution, elimination and Guassian elimination.
To solve a system of linear equations using graphs, each equation is plotted. The point of intersection would be the solution to the system. For a system of three variables a 3 dimensional plot can be used to find the intersection of the planes. At 4 or more variables it gets more complicated and other methods should be used.
The graph is plotted in 2 dimensions for the x and y variables.
In math, a system might have 3 dimensions or higher. The graph below is plotted in 3 dimensions for the x, y and z variables.
Substitution
Substitution works by solving for one of the variables in one equation and substituting it into the other equation. The second equation is solved for the remaining variable. This solution is then substituted into the other equation to test the answers. Example:
| 4x+2y=8 | |
| 3x+5y=6 | |
| Solve for y in the first equation: | |
| 2y = 8 - 4x | |
| y = 4 - 2x | |
| Substitute the expression for y which is 4 - 2x for y in the second equation: | |
| 3x + 5(4-2x) = 6 | |
| Solve for x: | |
| 3x + 20 - 10x = 6 | |
| 20 - 7x = 6 | |
| -7x = 6 - 20 | |
| x = 2 | |
| Substitute the value for x which is 2 back into the first equation to solve for y: | |
| 4(2) + 2y = 8 | |
| 8 + 2y = 8 | |
| y = 0 | |
The solutions can be checked by substituting both back into the system of equations.
| First equation: | |
| 4*2 + 2*0 = 8 | |
| 8 + 0 = 8 True | |
| Second equation: | |
| 3*2 + 5*0 = 6 | |
| 6 + 0 = 6 True |
Elimination
The method of elimination involves setting the coefficients of one variable equal to each other then subtracting one equation from another. This is done by multiplying both sides of each equation by a factor that will produce the LCM or Least Common Multiple. In some cases the situation is simplified when the other coefficient is already a multiple. Only one equation then has to be multiplied for a system of two equations. When the coefficients of the variable to be eliminated are equal then the equations are subtracted from each other. The equations can be added also if the coefficients are equal in magnitude but opposite in sign. Example:
| 5x+2y=10 | |
| 3x+y =7 | |
| The LCM is 2. Multiply the 2nd equation by 2: | |
| 5x+2y=10 | |
| 2(3x+y) = 2*7 | |
| The result should be: | |
| 5x+2y = 10 | |
| 6x+2y = 14 | |
| Next, subtract the second equation from the first: | |
| 5x+2y = 10 | |
| - 6x+2y = 14 | |
| _____________ | |
| -x + 0 = -4 | |
| The answer for the x is 4: | |
| x = 4 |
If there is one remaining variable, the values of the solved variables can be substituted back in or the other variables can be eliminated.
| Substitute the 4 back into the x of the first equation and solve for y: | |
| 5*4 + 2y = 10 | |
| 20 + 2y = 10 | |
| 2y = -10 | |
| y = -5 |
Now that both variables are known, substitute both into the second equation to see if the answers are true:
| 3*4 + (-5) = 7 | |
| 12 - 5 = 7 True |
Also, the method of elimination can be used on the other variable:
| Find the LCM for the x coefficients 5 and 3 in: | |
| 5x+2y=10 | |
| 3x+y =7 |
Since 3 and 5 are separate primes, the LCM is 5*3 = 15. The x coefficients should be equal to fifteen. Multiply the first equation be 3 and the 2nd equation by 5 to make the x coefficients equal:
| 3(5x+2y)=3*10 | |
| 5(3x+y) =5*7 | |
| The result should be: | |
| 15x+6y = 30 | |
| 15x+5y = 35 | |
| Next, subtract the second equation from the first: | |
| 15x+6y = 30 | |
| - 15x+5y = 35 | |
| ______________ | |
| 0 + y = -5 |
The answer for the second elimination is y = -5. This also resulted from substitution of x = 4.
Matrices
A system of linear equations can be represented by a matrix. It is easier to solve a large system with matrices than with the other methods presented here. Find out more about using a matrix to solve problems.
Matrix Algebra