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We have this system of equations here.

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3t plus 4g is equal to 6.

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And we have negative 6t plus g is also equal to 6.

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And there's a bunch of ways you can

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solve systems of equations.

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You can graph them.

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You can solve by substitution.

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Whenever you see the coefficients on one of the

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terms, in this case the t term, they're almost

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cancelable.

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And when I say cancelable, if I add 3t to negative 6t, you

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won't be able to cancel them out.

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The t won't disappear.

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But if I were to scale the 3 up, if I were to multiply it

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by 2 and it becomes 6t, and then I were to add these two

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things, then they would cancel out.

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And I would be left just with g's.

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So let's try to do that.

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And now remember, I can't just multiply this 3t by 2.

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In order for this equation to hold true, anything I do to

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the left-hand side, I have to do to the right-hand side as

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well, and I have to do it to the entire left-hand side.

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So let me multiply this equation by 2.

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So I'm multiplying it by 2.

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So let me just write it here.

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So I'm going to multiply 2 times 3t plus 4g is

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equal to 2 times 6.

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Anything I do to one side, I have to do to the other side.

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The equality still holds true.

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So 2 times 3t is 6t plus 2 times 4g is 8g, is equal to 2

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times 6, which is 12.

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So I really just rewrote the same information, the same

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constraint in this first equation.

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I just multiplied both sides by 2.

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Now, let me write the second equation right below it.

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I'll write it in orange.

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So the second equation right here is negative 6t plus g is

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equal to 6.

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Now, think about what happens if I were to

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add these two equations.

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Remember, I can do that because I'm essentially adding

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the same thing to both sides of this top equation.

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Or you could say I'm adding the same thing to both sides

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of this bottom equation because the other equation is

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an equality.

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This negative 6t plus g, it is 6.

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So if I'm adding 6 to 12, I'm really adding the same

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quantity to the left-hand side.

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That's why I can do it.

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So let's add the left-hand sides together.

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When we do that, the 6t's cancel out.

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That was the whole point behind multiplying this first

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character by 2, so that the 3t becomes a 6t, and we're left

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with 8g plus g is 9g, is equal to 12 plus 6, which is 18.

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Divide both sides by 9, and you are left with g is equal

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to 18/9, or 2.

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So we've solved for g.

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Now we can substitute back and solve for t, and we can use

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either of these equations.

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Let's use the second equation right here.

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So we have negative 6t plus g-- we just solved for g; g is

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2-- is equal to 6.

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And we can subtract 2 from both sides of this equation.

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Subtracting 2, the left-hand side of the equation, that

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cancels out.

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You have negative 6t is equal to 6 minus 2 is 4.

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Now I can divide both sides of this equation by negative 6.

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And we're left with t is equal to-- what is this?

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t is equal to negative 2/3.

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And we're done.

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We've solved for a t and a g that satisfy both equations.

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We just saw that it satisfies the bottom one.

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If you want to feel good that it satisfies the top,

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substitute them back into the top.

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Let's do that.

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3 times the t we got, negative 2/3, plus 4 times the g we

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got, so plus 4 times 2.

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Let's see what that is.

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3 times negative 2/3, that's negative 2.

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The 3's cancel out.

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Plus 4 times 2 is 8.

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Negative 2 plus 8 is equal to 6.

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And that's exactly what that first equation got us.

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So these two values definitely satisfy both equations.

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