Solving Systems of Linear Equations by Substitution and Elimination
Substitution and Elimination for Solving Linear Systems
Solve the system of equations using substitution and elimination. The system is. x plus 3y equals 5, 2x minus 4y equals 3. We’re going to start with substitution. To use substitution, or to substitute, means to replace with something else. To use substitution we’re going to replace one of our variables with the expression for one of our variables in the other equation, or for the same variable in the other equation.
Looking at my two equations in my system, it looks like it would be easiest to solve for x in the first equation. That’s what I’m gonna start with, x plus 3y, equals 5. To solve for x, we need to subtract 3y from both sides. These cancel and we get x equals -these are not like terms that cannot be added together- so -3y plus 5. Now that we know what x equals in the first equation, we can substitute this expression, -3y plus 5, for x in the second equation.
I have 2, 2 times x, and instead of x we’re replacing that with -3y, plus 5, minus 4y, equals 3. What we’ve done by substituting this expression -3y plus 5 for x, is we’ve gotten rid of x. Now, we only have 1 variable, y. Which means we can solve for y. To solve for y, we first need to distribute 2 times -3y is -6y, 2 times 5 is 10, minus 4y, equals 3.
Then we need to combine like terms, -6y and -4y. That’s -10y, plus 10, is equal to 3. Then subtract 10 from both sides. Those are additive inverses so they cancel. We have negative 10y is equal to 3 minus 10, is -7. Then finally we divide both sides by -10 to solve for y. y equals -7 divided by -10 is 7/10. Now that we know what y is, we can use that to substitute, we can substitute 7/10 for y, in either one of our equations, so that we can find x.
Really, the easiest equation to substituted it into, would be this one right here: x equals -3y plus 5. This is just our first equation rearranged. As you can see it’s already got x equals right there, so we just need to plug in that 7/10 for y and solve. x equals -3 times 7/10, plus 5. To multiply, I’m going to put -3/1, x equals -3 times 7 is -21, divided by 1 times 10 is 10, plus 5.
Now in order to add -21/10 to 5, we need to have a common denominator. 5 is 5/1, the least common denominator for 1 and 10 would be 10. I’m going to multiply numerator and denominator times 10. x equals -21/10 plus, 5 times 10 is 50, divided by 10. Now we just our numerators. x is equal to -21 plus 50 is 29, divided by 10. Or we could write it as a mixed number. 10 goes into 29 2 times with 9 leftover, 2 and 9/10.
When you’re asked to solve a system, the solution of a system is the ordered pair that satisfies both equations. It’s the ordered pair, or the point on the graph, where those two equations intersect. The answer to this problem is our ordered pair. x is 2 9/10, and y is 7/10. You could write it as a decimal, or in fraction form. That’s substitution.
You can tell by the answer, that finding this solution would have been pretty difficult if we were just graphing these systems on graph paper. We probably wouldn’t come up with this answer if we graphed it by hand on graph paper. Having these methods like substitution and elimination are very helpful for these kinds of answers. Elimination, to eliminate something means to get rid of it, and that’s what we want to do.
We have two variables in both of our equations, and with two different variables you can’t solve. We want to eliminate one of our variables. The first thing we’re going to do is stack our equations on top of each other. x plus 3y equals 5, and 2x minus 4y, equals 3. For elimination, what we’re doing is adding the two equations together to eliminate one of the variables. As they are now, if we added these two equations together neither one of the variables would eliminate.
x plus 2x is 3x, 3y and -4y would just give us -y. We can change these equations by multiplying them times anything we want. As long as we’re consistent and we multiply every term in that equation by the same number, we can multiply them by anything we want. In order to eliminate x, this would need to be a -2x. I’m just going to multiply this entire equation -2.
Every term has to be multiplied by -2. -2 times x, negative 2x. -2 times 3y, -6y. Then -2 times 5, -10. It’s very important that every term get multiplied by the same number, so that the equation is still balanced and equal. I’m going to run a line through this one so it doesn’t distract us. Now we’re going to add our two equations together. 2x and -2x is 0. -4y and -6y is -10y. That equals 3 and -10 is -7.
Now to solve for y, we just divide both sides by -10. Those cancel, and y is equal to, a negative divided by a negative is a positive, 7/10 again. It’s the same thing we got when using substitution. The second part of elimination is finding x. To find x you actually use substitution. We’re going to substitute y equals 7/10 into either one of our equations.
I’m going to use that first equation again, just because it looks a little bit easier to use. x plus 3y equals 5. y is 7/10. x plus 3 times 7/10, equals 5. x plus -if I’m going to multiply those together, put 3/1, multiply across, 3 times 7 is 21, 1 times 10 is 10. 21/ 10, and that’s equal to 5. Then to solve for x, we need to subtract the 21/10 from both sides. Those are inverses, so they cancel. x equals 5 minus 21/10.
Again, we need a common denominator in order to subtract. This is 5/1, and we’re gonna multiply numerator and denominator times 10. This is 50/10 minus 21/10. x, again, is equal to 29/10. Which is, again, 2 9/10. Finally, our answer is our ordered pair. x is 2 9/10, and y is 7/10. That’s our solution found using elimination.