Transposing is an important technique for students to master as they progress from Algebra I to Algebra II and beyond. It’s the process of moving quantities across the equal sign in an equation to make the problem solving process more efficient. Here’s how it works.
Solving an equation means isolating the variable on one side of the equation, either the right side or the left. Generally it is better to isolate the variable on the left side of the equal sign because we read from left to right.
Beginning algebra students are taught the Law of Equations: whatever you do on one side of an equation, you must do on the other side as well.
Using the Law of Equations, to solve 2x – 3 = 13, we can add 3 to both sides, giving us:
2x – 3 + 3 = 13 + 3 ===>2x = 16
Then, we divide both sides by 2 to find that x = 8.
To solve 2x –3 = 13 by transposing, we do the following:
Instead of adding 3 to both sides,we can transpose the 3 that’s already there by
The beauty of this process is that it is visuallyeasier to see how the parts of the equationchange as we solve it. Instead of writing the transposed quantity twice (once on the left and once on the right), we literally pick up the quantity and move it to the other side. The price we have to pay for doing this is changing the sign of the quantity being moved.
Let’s solve 3y + 2x – 3 = 7 for y.
Since we want to isolate y, we can transpose 2xand – 3.
This gives us y = –2x + 7 + 3. Simplifying, we get y = –2x + 10.
If we used the Law of Equations, we would have to write out
3y + 2x – 3 = 7 ===> 3y + 2x – 2x – 3 + 3 = 7 – 2x + 3
While simplifyingthe above does give us y = –2x + 10, it comes with a lot more work.
Note that most classroom teachers will not allow transposing until students learn to use the Law of Equations.
Try these:
-
Solve for x: 4x + 9 = 49
-
Solve for y: 2y + 7 – 3 = 24
-
Solve for y: 3x + 9y – 7 = 49
-
Solve for a:3b – 4a – 6 +4 = 49
-
Solve for a:2b + 4a + 5 – 2 = 42 – 2a
