Systems of Linear Equations: Solving by Addition / Elimination

When I was solving one-variable linear equations, back in the day, I would "cancel out" an unwanted number by adding its opposite. (In the example above, this would have been the −6 that was added in the second line, in order to cancel out the +6.) Then I'd draw a horizontal "equals" line under what I'd added to both sides of the original equation, and I'd add down. This would get the variable by itself on one side of the "equals" sign.

I want to do something similar here. I know how to solve linear equations with one variable. Here I've got two. Can I get rid of one of these variables in the system, just as I'd have gotten rid of the −6 in the equation?

Looking at the system of equations they've given me, I see that I've got a +y in the first line, and a −y in the second line. If I added these, they'd cancel out, leaving me with just the variable x. In other words, if I add down, I should end up with a linear equation with just one variable, and I know how to solve those. So let's do that!

I write down the two equations, draw an "equals" bar under them, and add down:

Now I have a one-variable linear equation that I already know how to solve. I divide through on both sides by 5 to get x = 5. This is half of the solution to this system.

(By the way, this adding of the two equations, or two "rows", is called a "row operation".)

To find the other half (that is, to find the y-value), I can plug this x-value back into either one of the original equations, and simplify for the value of y. (This process — of taking a partial solution and plugging it back in to some portion of the original exercise to find the rest of the solution — is called "back-solving".)

I can use either of the original equations to back-solve and find the value of y. The first equation has smaller numbers (and I'm lazy), so I'll back-solve in that one:

This gives me the other half of the solution, so my answer is:

3(5) − y = 16
  15 − y = 16
        −y = 1
          y = −1

Note that the x-terms would cancel out if only they'd had opposite signs. But I can create this opposite-sign cancellation by multiplying either one of the equations by −1, and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the −1 through the entire equation. (That means both sides of the "equals" sign!)

I flipped a coin; I'll multiply the second equation.

(The "−1R₂" notation over the arrow in the above image indicates that I multiplied row 2 by −1. This "R_n" notation, indicating that you're doing something with the n-th row, is standard. And this multiplying of a row by a numerical value is another "row operation".)

By setting up the x-terms to cancel out when the equations are added together, I have eliminated that variable. Now I can solve the resulting one-variable equation "−5y = −25" to get y = 5.

To find the corresponding value of x, I plug this y-value back into either of the original equations. Back-solving in the first equation, I get:

This gives me the other coordinate of the solution point, so my answer is:

Nothing cancels here, but I can multiply to create a cancellation. (As long as I multiply both sides of the equation by the same value, I won't have changed anything in mathematical terms. But I may be able to change things in practical terms, to create a cancellation.) If I multiply the first equation by 4, this will set up the y-terms to cancel.

Solving this, I get that x = 2. I'll use the first equation for backsolving, because the coefficients are smaller (and I'm lazy).

Now I have the two coordinates of the solution point:

Hmm... As the system stands, nothing cancels. But I know that I can multiply to create a cancellation.

In this case, neither variable is an obvious choice for cancellation, so I'll consider the least common multiples of the coefficients. I can multiply the equations (by 3 and 4, respectively) to convert the x-terms to 12x's, or I can multiply them (by 8 and 3, respectively) to convert the y-terms to 24y's. Since I'm lazy and 12 is smaller than 24, I'll multiply to cancel the x-terms.

(I would get the same answer in the end if I set up the y-terms to cancel. It's not that how I'm doing it is "the right way"; it was just my choice. You could make a different choice, and your choice would be just as correct as mine.)

I will multiply the first row by 3 and the second row by 4; then I'll add down and solve.

Solving, I get that y = 5. Neither equation looks particularly better than the other for back-solving, so I'll flip a coin and use the first equation.

Remembering to put the x-coordinate first in the solution, I get:

I think I'll multiply the second equation by 2; this will at least get rid of the decimal place.

Oops! This result isn't true! Zero is never equal to −2!

All of my steps were correct, but I ended up with garbage. This tells me that my original assumption (being that the system had a solution) must have been wrong. So this is an inconsistent system (that i s, one that graphs as two parallel lines) with no solution (that is, having no intersection point).

I think it'll be simplest to cancel off the y-terms, so I'll multiply the second row by −3.

Well, yes, zero does equal zero, but...?

I already knew that zero equals zero. This information doesn't add anything to my store of knowledge. In particular, it doesn't help me narrow down my answer to one solution point. All my math was correct, so the issue lies elsewhere.

Then I remember: If the two equations are really the same one equation, then this "zero equals zero" result is the sort of thing I should expect. In fact, this result tells me that this system is a dependent system (that is, one that graphs as just one line) and, solving either of the original equations for "y=", I find that the solution is the equation of the whole line, namely: