Algebra I
Linear Systems: Solve by Graphing & Substitution
Big Picture
To find the solutions for systems of linear equations, the graphing method or substitution may be employed. However, not every system has exactly one solution. Some systems may have more than one solution and some may have no solutions at all.
Key Terms
System of Linear Equations: A set of equations that must be solved together to find the solution that fits them both.
Consistent System: A linear system that has exactly one solution.
Inconsistent System: A linear system that has no solution.
Dependent System: A linear system with an infinite number of solutions.
Systems of Linear Equations
A system of linear equations (also called a linear system) is two or more linear equations containing the same variables. The solution to a linear system must be a solution for ALL of the equations.
Solve by Graphing
When graphed, the solution of the linear system is the point where all the lines intersect (cross).
Example:
- Point \(A\) is not on either of the lines, so it is not a solution to the system.
- Point \(B\) is only on one of the lines, so it is not a solution to the system.
- Point \(C\) is on both lines, so it is the solution to the system. It is the only point that is on both lines, so it is the only solution.
To solve by graphing:
- Graph the two equations. The solution is the intersection of the two lines.
- Check the solution by plugging it into both equations.
- Make sure the solution works for both equations!
It is usually easier to see the solution on a graph, but this method can lead to imprecise answers if the solution is anon-integer.
Solve by Substitution
The substitution method is a way to solve for the solution more precisely.
- For one of the equations, isolate one of the variables so that there is only one variable on one side of the equation.
- Plug the expression in step 1 and solve for the other variable.
- Plug the value in step 2 into the first equation and solve.
Example: \(2x + 3y = 6\) and \(-4x + y = 2\)
1. Isolate \(y\) in the second equation \(y = 2 + 4x\)
3. Plug the value for \(x\) into the first equation and solve for \(y\).
\(y = 2 + 4(0)\)
\(y = 2\)
2. Plug the expression into the first equation and solve for \(x\).
\(2x + 3(2 + 4x) = 6\)
\(2x + 6 + 12x = 6\)
\(14x = 0\)
\(x = 0\)
The solution is \((0, 2)\). Check by plugging it into both equations.
\(2(0)+3(2) = 6\)
\(-4(0)+2 = 2\)
Both equations are true, so \((0, 2)\) is the solution.
This method works on all systems, but we usually need to do extra work to get the equations into the form that we want.
