Expand menu menu_open Minimize Start chapters Home History history History expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
search
menu_open
{{ courseTrack.displayTitle }}
{{ statistics.percent }}% Sign in to view progress
{{ printedBook.courseTrack.name }} {{ printedBook.name }}
search Use offline Tools apps
Login account_circle menu_open

Solving Linear Systems

Solving Linear Systems 1.13 - Solution

arrow_back Return to Solving Linear Systems
a

We want to solve the following system of linear equations by substitution. When solving a system of equations using substitution, there are three steps.

  1. Isolate a variable in one of the equations.
  2. Substitute the expression for that variable into the other equation and solve.
  3. Substitute this solution into one of the equations and solve for the value of the other variable.
For this exercise, is already isolated in Equation I. Let's substitute this into Equation II, and solve the resulting equation for
To find the value of we need to substitute into either one of the equations in the given system. Let's use the first equation.
The solution, or point of intersection, to this system of equations is the point
b
Here we want to solve the following system of linear equations. Let's start by isolating in the equation II. Now that we've isolated we can substitute its value into and solve the resulting equation for
The solution, or point of intersection, to this system of equations is the point