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Solving Systems of Equations using Elimination - Problem 3 7,319 views
Here I have a difficult system of equations and I'm asked to solve it using elimination. I know it's difficult because my coefficients for x and my coefficients for Ys don't look at all like additive inverses.
So before I can solve this system I'm going to have to do some multiplying both on the top equation and the bottom equation. But I want to make sure I multiply in a clever way so that my coefficients would cancel out.
You can choose which variable you want to eliminate I'm going to choose to eliminate Xs, I just made that up you could you could also choose to eliminate Ys. In order to eliminate my Xs I will need this number to be additive inverses. So here's a trick I'm going to multiply this top equation by -5 which is like that coefficient, I'm going to multiply the second equation by +4 because when I do that distributing my x coefficient here will be -20, here I would have +20 and those are additive inverses.
The trickiest thing of this whole problem is what I just did. I used the coefficient from the bottom to multiply by the top and I used the coefficient from the top to multiply the bottom one by. I also sneaked in there the negative sign because I need to have a -20 and a positive 20 or a negative value and a positive value in order fro them to be additive inverses.
Once you have it set up this is just straight forward plugging and chugging algebra solving skills. The first equation I'll have -20x plus 15y, careful with the minus signs, is equal to positive what's 313 times 5, I'm not totally sure let's see 5 carry the 1 is 65 okay. So I'm going to have +65 over here then on the second equation I'm going to have 20x plus 8y is equal to 4. Now I'm ready to eliminate. This equation is equal to that equation I just multiplied every term by -5 I didn't change the value of anything I just multiplied the coefficients.
That's mathematically allowed, it's like I didn't change any of the outcomes. Now I can eliminate by adding vertically my x is eliminated I have 23y is equal to 69 and if you divide both sides by 23 you end up getting that y is equal to 3. That's half of my solution to find my x umber I'm going to go back to the very original equation just I don't want to use one of these guys in case I made a mistake. Let's go back to the very first equation and find x by substituting in 3 from my y value and solve for x. 4x take away 9 is equal to -13. Add 9 to both sides so that 4x is equal to -4, x is equal to -1.
Good, I think I'm done the last thing I want to do is be sure I check my work. I did all kinds of Mathy stiff in here that might have been an error so in order to check your work, make sure you plug in your xy coordinates into both original equations and make sure you get equalities.
4 times my x number take away 3 times 1 number should equal -13. Good it does, that's my first equation let's get the second one. 5 times my x number times 2 times my y number should be 1. Oops! +1 not negative, +1. -5 plus 6 good it's equal to 1. That tells me I did the problem correctly.
Again the hardest step with this problem is the very first step and that was where I looked in my coefficients I saw that they were not additive inverses. So I chose to eliminate x looking at the coefficients of x, you can also choose y and then multiply this top equation by 5, multiply the bottom equation by 4 and then I went back and made one of them have a negative sign. That way I had -20 and +20 which are my additive inverse coefficients.