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

CCSS.Math:

## Video transcript

we have this system of equations here 3 T plus 4 G is equal to 6 and we have negative 6 T 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 but whenever you see the coefficients on on one of the terms in this case the T term and they're all there 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 60 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 now remember I can't just multiply this 3 T by 2 in order for this equation to hold true anything I do the left hand side I have to do to the right hand side as well so let have to do to the entire left hand side so let me multiply 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 4 G is equal to 2 times 6 anything I do is one side I have to do the other side the Equality still holds true so 2 times 3t is 60 plus 2 times 4 G is 8 G 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 and I'll write it in orange so the second equation right here is negative 6 T plus G is equal to 6 now think about what happens if I were to add these two equations and 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 inequality this negative 6t plus G it's 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 60s cancel out that was the whole point behind multiplying this first character by 2 so that the 3t becomes the 60 and we're left with 8 G Plus G is 9 G is equal to 12 plus 6 which is 18 divide both sides by 9 and you are left with G is equal to 18 over 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 to is equal to is equal to 6 and we can subtract 2 from both sides of this equation subtracting to 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 negative 6 and we're left with T is equal to what is this negative T is equal to negative 2/3 and we're done we've solved for a T and a G that satisfy both the 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 3 times 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 3s 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