1. G = B + 1
3B + G = 5

3. You have learned to solve a one-variable equation algebraically.
In this lesson, we will apply that knowledge to solving pairs of linear equations.

4. A solution to a pair of linear equations is a pair of values that makes both equations
true at the same time.
A pair of linear equations is sometimes called a linear system.

5. Students sometimes think that they must solve for the variables in a particular order, such as alphabetically. Because the value of each variable stays the same for each equation, the problems can be solved in any order.
However, it is most convenient to substitute an expression for a variable that is already isolated, if there is one.

6. From the second equation, we know that G = 3.
We can swap the G for 3 in the first equation since every G in this system equals 3.
Now we have enough information to find out what B is. We can solve this one by inspection. 3 + what = 8? B is 5.

7. G + B = 8
(3) + (5) = 8
8 = 8
G = 3
(3) = 3
Let’s algebraically check our work to make SURE it is a solution that BOTH equations have in common. It’s a solution.
We can write our solution as an ordered pair (B, G) = (5, 3)

8. Both of these equations have G already isolated
Both of these equations have G already isolated. Since G is B + 4, we can swap G for B + 4.

9. Once we have substituted B + 4 for G into the second equation, let’s concentrate on the bottom equation since it only has one variable.

10. For any number B, the value of the left expressions equals the value of the left expression. Subtracting 2 still means that the number of pencils on each side are still equal, because both sides are just two less than they were before.
Can we do the same with removing variables from both sides of the equation? Yes!
B = 2.

11. Now that we know B = 2, we can go back to the first equation and see that G = 6.

12. G = B + 4 (6) = (2) + 4 6 = 6 G = 2B + 2 (6) = 2(2) + 2 6 = 4 + 2
We need to check that this solution is in common with BOTH equations.

13. In this example, G is equal to one more than B
In this example, G is equal to one more than B. We can use this information to SWAP the G for B + 1 in the second equation. By doing this, we now only need to deal with only the variable B.
Let’s focus on the bottom equation.

14. This equation can be simplified by combining like terms.
We can now begin to isolate B by subtracting 1 from both sides.
Dividing by 4 shows us that B = 1.

15. B = 1 and G = 2.
Now that we have solved B = 1, we can use that information and see that G = 2.

16. 3B + G = 5 G = 1 + B 3(1) + 2 = 5 (2) = (1) + 1 3 + 2 = 5 2 = 2 5 = 5
Algebraic check is final step.