# Systems of equations with elimination: 3t+4g=6 &Â -6t+g=6

CCSS Math: 8.EE.C.8b

## Video transcript

We have this system of equations here. 3t plus 4g is equal to 6. And we have negative 6t plus g is also equal to 6. And there's a bunch of ways you can solve systems of equations. You can graph them. You can solve by substitution. Whenever you see the coefficients on one of the terms, in this case the t term, they're almost cancelable. And when I say cancelable, if I add 3t to negative 6t, you won't be able to cancel them out. The t won't disappear. But if I were to scale the 3 up, if I were to multiply it by 2 and it becomes 6t, and then I were to add these two things, then they would cancel out. And I would be left just with g's. So let's try to do that. And now remember, I can't just multiply this 3t by 2. In order for this equation to hold true, anything I do to the left-hand side, I have to do to the right-hand side as well, and I have to do it to the entire left-hand side. So let me multiply this equation by 2. So I'm multiplying it by 2. So let me just write it here. So I'm going to multiply 2 times 3t plus 4g is equal to 2 times 6. Anything I do to one side, I have to do to the other side. The equality still holds true. So 2 times 3t is 6t plus 2 times 4g is 8g, is equal to 2 times 6, which is 12. So I really just rewrote the same information, the same constraint in this first equation. I just multiplied both sides by 2. Now, let me write the second equation right below it. I'll write it in orange. So the second equation right here is negative 6t plus g is equal to 6. Now, think about what happens if I were to add these two equations. Remember, I can do that because I'm essentially adding the same thing to both sides of this top equation. Or you could say I'm adding the same thing to both sides of this bottom equation because the other equation is an equality. This negative 6t plus g, it is 6. So if I'm adding 6 to 12, I'm really adding the same quantity to the left-hand side. That's why I can do it. So let's add the left-hand sides together. When we do that, the 6t's cancel out. That was the whole point behind multiplying this first character by 2, so that the 3t becomes a 6t, and we're left with 8g plus g is 9g, is equal to 12 plus 6, which is 18. Divide both sides by 9, and you are left with g is equal to 18/9, or 2. So we've solved for g. Now we can substitute back and solve for t, and we can use either of these equations. Let's use the second equation right here. So we have negative 6t plus g-- we just solved for g; g is 2-- is equal to 6. And we can subtract 2 from both sides of this equation. Subtracting 2, the left-hand side of the equation, that cancels out. You have negative 6t is equal to 6 minus 2 is 4. Now I can divide both sides of this equation by negative 6. And we're left with t is equal to-- what is this? t is equal to negative 2/3. And we're done. We've solved for a t and a g that satisfy both equations. We just saw that it satisfies the bottom one. If you want to feel good that it satisfies the top, substitute them back into the top. Let's do that. 3 times the t we got, negative 2/3, plus 4 times the g we got, so plus 4 times 2. Let's see what that is. 3 times negative 2/3, that's negative 2. The 3's cancel out. Plus 4 times 2 is 8. Negative 2 plus 8 is equal to 6. And that's exactly what that first equation got us. So these two values definitely satisfy both equations.