MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
In a system of equations, if one variable is already isolated (i.e., one variable is already solved for), you can easily use substitution to find the solution. If one equation is solved for x, then take the value of x and plug it into the other equation. Solve for y, the unknown variable, then plug that value into one of the original equations to find the value of x. The x and y values make up the coordinates of the solution to the system. Check your answer by plugging the values into both equations. If the equations are true, your solution is correct.
Here is a system of equations that I want to solve using substitution and I’m already set up pretty well because x is already isolated. I know that x is equals to the expression 4y-2, so when I’m solving this problem in the second equation instead of writing x right there, I’m going to write parentheses 4y minus 2. Here is what I mean 2 times what used to be x, but now I’m calling 4y minus 2 plus my y value, is equal to 14.
From there I can solve this problem pretty quickly, pretty easily; it’s just using distributing, simplifying and solving. 8y minus 4 plus y equals 14, combine those so I’ll have 9y minus 4 equals 14. Add 4 to both sides, y is equal to 2. That’s only half my answer, though don’t start having the part yet, we’re not all the way there. We still need to find the x co-ordinate that goes along into my solution.
In order to find the x coordinate, I’m going to plug in my y value of 2 into either original equation, either one should give me the same x answer, so let me come over here so I have a little more space and I’m going to plug in 2 for that y value right there. X is equal to 4× times 2 now take away 2, 8 take away 2 so my x value I think is 6. Before I move on I want to make sure I check my answer, and the way I can check is plugging this ordered pair, into both original equations. It has to work in both in order for it to be a system of equations solution.
So here comes my check. In the first equation, is it true that my x value is equal to 4 times my y value take away 2? Let’s see 6 equals 8 minus 2 yeah that works, so that first equation works that’s why I’m halfway there with my checking.
Now I need to do the second equation, 2 times my x value plus my y value should be equal to 14. Good since both of these gave me true equalities that means this point is indeed the solution. This is where these two lines would cross if I graphed them, but I did it without having to graph which is great for me because I don’t like graphing.
So again you guys I looked at this problem and I knew substitution was going to work well for me because one of variables was already isolated. Then I just substituted this expression into the other equation so I had one equation with one variable to solve.
