#### darkside55

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If these two triangles are congruent, what two pairs of sides must be equal? What two equations can you write?

Then solve the resulting system of equations for x and y.

If you have attempted some work, please show it to us, so we can see how close you are to getting it right, and encourage you!

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Hello lev888, I have tried to set up the equations two different ways both ending with different answers the first time I had done this I had set up the equations as: x=y+1 and 3y=x+3 so what I tried to do was solve for Y in the second part and what i got was that Y=x because I had divided by three on both sides which is where I had started to get confused. The other equation I had done was 2x+3=4y+1(i just combined my like terms together) then i had subtracted the 1 from the positive three to get 2+2x=4y then I had divided the entire equation by 2 and i got x=2y. Yes i do understand all the terms. I reviewed them for the past two hours and i still am as confused as beforeWhat have you tried? Do you understand all the terms? Did you review the relevant chapter(s)?

thanks~ elyse

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Hello Dr. Peterson the HL theorem is when right triangles have a congruent hypotenuse, along with a congruent leg the sides that would be equal would be x and y+1 along with x+3 and 3y for the first one it would still be x=y+1 and the second im pretty sure would be 1/3x=y but even with these answers I am still baffled by it and still curious if I might have missed a step or done something wrong

If these two triangles are congruent, what two pairs of sides must be equal? What two equations can you write?

Then solve the resulting system of equations for x and y.

If you have attempted some work, please show it to us, so we can see how close you are to getting it right, and encourage you!

thanks~elyse

Ok, you have this system: x=y+1 and 3y=x+3Hello lev888, I have tried to set up the equations two different ways both ending with different answers the first time I had done this I had set up the equations as: x=y+1 and 3y=x+3 so what I tried to do was solve for Y in the second part and what i got was that Y=x because I had divided by three on both sides which is where I had started to get confused. The other equation I had done was 2x+3=4y+1(i just combined my like terms together) then i had subtracted the 1 from the positive three to get 2+2x=4y then I had divided the entire equation by 2 and i got x=2y. Yes i do understand all the terms. I reviewed them for the past two hours and i still am as confused as before

thanks~ elyse

I am not sure how you divided the 2nd equation by 3 and got y=x. If you divide x+3 by 3 you should get x/3 + 1. Please look into this.

Here's how I would start. The first equation already gives you x expressed through y. So use it to substitute (y+1) for x in the second equation and then solve it for y.

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No ...Note the following. 25 = 15 + 5 OK?

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Ah, so my whole point is void.No ...

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Note the following. 20 = 15 + 5 OK?

Now we divide both sides by 5. So 20/5 = 15 +5/5. Then 4 = 15 + 1 which is not true.

On the other hand, 20/5 = 15/5 +5/5. Then 4 = 3 + 1 which is true.

You need to divide all terms by 5 or 4 or whatever you want.

So when you divide x+3 by 3 you need to get x/3 + 3/3 =x/3 + 1

Also 3/3 is not 0, but rather 1, so you can't ignore it!

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Well, not entirely. Valid logic can be applied to a false premise ...Ah, so my whole point is void.