Advanced Vector Operations and Applications: Background You’ll Need 2
Use substitution and the addition/elimination method to solve systems of equations with two variables
Solving Systems of Equations by Substitution
When solving systems of linear equations, we need methods that work reliably for all types of solutions—whether they involve integers, decimals, or fractions. The substitution method provides an algebraic approach that gives us exact answers.
The substitution method works in two main steps: solve one equation for one variable, then substitute that expression into the other equation. This allows us to solve for one variable at a time, which simplifies the process and ensures accuracy.
How To: Given a system of two equations in two variables, solve using the substitution method.
Solve one of the two equations for one of the variables in terms of the other.
Substitute the expression for this variable into the second equation, then solve for the remaining variable.
Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
Check the solution in both equations.
Solve the following system of equations by substitution.
Clearly, this statement is a contradiction because [latex]9\ne 13[/latex]. Therefore, the system has no solution.
Graph demonstrating an inconsistent system
Analysis
Let’s graph the equations to confirm that the system has no solution. Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common.
Solving Systems of Equations by the Addition Method
Another method of solving systems of linear equations is the addition method, this method is also called the elimination method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.
How To: Given a system of equations, solve using the addition method.
Write both equations with [latex]x[/latex]– and [latex]y[/latex]-variables on the left side of the equal sign and constants on the right.
Write one equation above the other, lining up corresponding variables.
If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable.
If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
Solve the resulting equation for the remaining variable.
Substitute that value into one of the original equations and solve for the second variable.
Check the solution by substituting the values into the other equation.
Both equations are already set equal to a constant. Notice that the coefficient of [latex]x[/latex] in the second equation, –1, is the opposite of the coefficient of [latex]x[/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[/latex] without needing to multiply by a constant.
A graph of two lines that intersect at the point negative seven-thirds, two-thirds
We gain an important perspective on systems of equations by looking at the graphical representation. See the graph to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.
Solve the given system of equations by the addition method.
[latex]\begin{align}3x+5y&=-11 \\ x - 2y&=11 \end{align}[/latex]
Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[/latex] in it and the second equation has [latex]x[/latex]. So if we multiply the second equation by [latex]-3,\text{}[/latex] the x-terms will add to zero.
[latex]\begin{align}x - 2y&=11 \\ -3\left(x - 2y\right)&=-3\left(11\right) && \text{Multiply both sides by }-3 \\ -3x+6y&=-33 && \text{Use the distributive property}. \end{align}[/latex]
