Find the X and Y-Intercepts of a Line Using Algebra

[One speaker.]

[The video opens to the title on top of a white background. The speaker uses the white background to work on problems and explain her solutions.]

Speaker: Welcome! In this video we will discuss how to find the X and Y intercepts of a line using Algebra. Here’s an equation. [Y equals ¾ X minus 2] This equation happens to be written in slope intercept form of a line. Let’s use it to help us algebraically the X intercept and the Y intercept. So the X intercept is found whenever y equals zero. In other words when y equals zero. And the Y intercept is found whenever X equals zero. So using this information, let’s find our X and Y intercepts.

So let’s start with the X intercept. So we find the X intercept by setting the Y intercept equal to zero and then solving for X. So the X intercept is found when zero is equal to ¾ X minus 2. [rewrites equation with zero substituted for Y] And now we solve for X. So first we’re going to add two to both sides. I’m going to add 2 on the right, and I’m going to add 2 to the left. [writes a plus 2 on both sides of the equation] And then I’m going to rewrite my equation. Now, on the left hand side we have 2 is equal to ¾ X, and then a negative 2 plus 2 is just zero. So we’re just left with the ¾ X on the right hand side. So now, in order to isolate X, we have to multiply by the multiplicative inverse of ¾ to both sides. We’re going to multiply everything on the right by 4/3, and multiply everything on the left by 4/3. [writes 4/3 on each side of the equation] ¾ times 4/3, we’re distributing this in, [points to 4/3] is the same as 1 because the 3 divided by 3 is 1 [crosses out the 3’s in ¾ and 4/3] and 4 divided by 4 is 1, [crosses out the 4’s] so all we’re left with is 1X, or in other words just plain X on the right hand side. That is equal to, remember 4/3 times 2 is the same as 4/3 times 2 divided by 1. [writes a 1 in the denominator under 2 on the left hand side of the equation] Now we can multiply the numerators and the denominators. 4 times 2 is 8 and 3 times 1 is 3. So X is 8/3 when Y is equal to zero. So our X intercept is at the coordinate 8/3, 0. Now, if we’re not too comfortable with graphing 8/3, we can change this into a mixed fraction or into a decimal. 8/3 is the same as 2 and ⅔. Let’s check that by following the rule to convert from a mixed fraction to an improper fraction. So you would take the whole number 2, multiply it by the denominator 3, 2 times 3 is 6 plus 2 more in the numerator is 8, and then put that all over the denominator which is 3. So 8/3, 2 times 3 6 plus 2 is 8 and then 3 is our denominator. So it is.

Now let’s go back and calculate for the Y intercept. The wonderful thing about this equation of a line is that the b, over here, [points to the negative 2 in the original equation] is the Y intercept. But our equation is actually y equals m x plus b. [writes equation] So if we have a negative 2 up here, [points to the negative 2 in the original equation] that’s the same as y equals ¾ X plus a negative 2. [writes equation] So in that case, b is equal to negative 2. So our Y intercept is at zero, negative 2. Let’s see why this works. Why does that happen? We know that the Y intercept is whenever X is equal to zero, so if we plug zero into this equation, we can solve for Y. Y equals ¾ times zero plus a negative 2. [rewrites equation with zero substituted for X] Well, ¾ times zero is just zero; anything times zero is zero. So we’re left with Y equals a negative 2. [writes down] And that’s why when our equation is in the from of Y equals m x plus b, the b value is the Y intercept value. So our Y intercept is at the point zero comma negative 2. When X is zero, Y is negative 2.

[The writing is cleared and a new equation appears on the white background.  The equation is 3 Y minus 6 X equals 12.]

In the previous example, the equation was in the slope intercept form of a line. Here’s another equation of a line, but this time it is not written in that format. But we can still use the same principle to help us find the X and Y intercepts. Again, the X intercept is found when Y equals zero, and the Y intercept is found when X equals zero. By plugging this knowledge into our equation, we can then solve for the other variable, in both cases, to find the X and Y intercept. So let’s solve for the X intercept first. So at the X intercept, our Y equals zero, so we’ll plug that into our equation. 3 times zero, since we’re finding the X intercept, minus 6 X equals 12. [rewrites equation with zero substituted in for Y] 3 times zero is just zero so we’re left with negative 6 X on the left hand side equal to 12. If we divide both sides by a negative 6, or, in other words, multiply by 1 over negative 6 to both sides of the equation, [writes a negative ⅙ on both sides of the equation] then our negative 6’s will cancel, [crosses out negative 6’s on the left hand side of equation] negative 6 divided by negative 6 is 1. We’re just left with 1 X on the left hand side equal to 12 divided by a negative 6 is negative 2. So our X intercept, when Y equals zero, is at the point negative 2 comma zero. Now let’s solve for Y intercept. The Y intercept occurs when X is zero, so let’s plug that into our original equation. [rewrites equation substituting zero for X and goes through the same process to find the Y intercept.] 3 Y minus 6 times zero equals 12. 6 times zero is zero, so all we’re left with is 3 Y equals 12. Divide both sides by 3, or in other words, multiply by one third on each side. 3 divided by 3 is 1 which leaves us with 1 Y equals 12 divided by 3 is 4. So our Y intercept is at the point zero comma 4. And that is how to use algebra to find the X and Y intercepts.

[End of Video.]