Video: Solving a System of Linear Equations

Consider the system of equations 3π‘₯ + 3𝑦 = βˆ’3, 3π‘₯ βˆ’ 2𝑦 = βˆ’8. Solve to find the value of 𝑦.

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Video Transcript

Consider the system of equations. Three π‘₯ plus three 𝑦 equals negative three. Three π‘₯ minus two 𝑦 equals negative eight. Solve to find the value of 𝑦.

When we’re considering systems of equations, solving means finding the place where these two lines would intersect. The place where their π‘₯-values and 𝑦-values are the same. Let’s look at two different ways to do this. First, I labeled the equations equation one and equation two. The first method we’ll consider is called solving by elimination. I want to consider the elimination method because I see that both of our equations have a three π‘₯-variable. So we’ll write the first equation three π‘₯ plus three 𝑦 equals negative three. And then, we’ll add the second equation directly below it. Three π‘₯ minus two 𝑦 equals negative eight. Making sure that we line up our π‘₯-variables, our 𝑦-variables, and what’s behind the equal sign. And then, we’re going to subtract the entire second equation from the first equation. That means we’ll subtract three π‘₯ from three π‘₯. Three π‘₯ minus three π‘₯ equals zero.

The next step is saying three 𝑦 minus negative two 𝑦. We need to be careful with our signs here. We are subtracting a negative, which means we need to add three 𝑦 and two 𝑦, which will give us positive five 𝑦. And then, we’ll have negative three minus negative eight. Again, we will be careful with our signs here. Negative three minus negative eight is the same thing as negative three plus eight. Negative three plus eight equals positive five. We now have a new equation that says five 𝑦 is equal to five. Our goal is to solve for 𝑦. And so we’ll divide both sides of the equation by five. Five 𝑦 divided by five is 𝑦. The fives cancel out. And five divided by five is one. Using the elimination method, we found that 𝑦 equals one.

Let’s consider another method. This time, we’ll solve this system using substitution. This time, we’ll take our first equation. Three π‘₯ plus three 𝑦 equals negative three. And we’ll solve for π‘₯. That means we’ll get π‘₯ by itself. Here, we notice that both the coefficients of π‘₯ and 𝑦 and the constant value are all multiples of three. And so we divide the entire equation by three, which would then give us π‘₯ plus 𝑦 equals negative one. Remember, our goal is to solve for π‘₯. And to do that, we’ll subtract 𝑦 from both sides. And that gives us the equation π‘₯ equals negative one minus 𝑦. We can call this equation three.

We now want to take equation three and substitute it back in to equation two. Equation two is three π‘₯ minus two 𝑦 equals negative eight. In place of π‘₯, we want to substitute negative one minus 𝑦. We then have three times negative one minus 𝑦 minus two 𝑦 equals negative eight. We need to distribute this three. Three times negative one is negative three. Three times negative 𝑦 is negative three 𝑦. And then we’ll bring down the rest of the equation. We can combine like terms. Negative three 𝑦 minus two 𝑦 equals negative five 𝑦. And everything else comes down. So we add three to both sides of the equation. And we have negative five 𝑦 equals negative five. Since our goal is to solve for 𝑦, we divide both sides of the equation by negative five. Negative five 𝑦 divided by negative five equals 𝑦. And negative five divided by negative five equals one. So both methods show that 𝑦 equals one.

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