Circles M and K are congruent, QR ≅ LN and OP ≅ VW . Find x and y.
Could you just rewrite one of the equations as variable = stuff then use that to stick into the other equation? I was thinking the easy way would be to take x+y=6 and make it x = 6-y. Then I would use the substitution method, plugging "6-y" into the other equation wherever there was an X. So, doing that would give us: 3(6-y) - y = 10. Now it only has ONE variable, so we can solve for Y, and once we have that we could use the known value for Y to solve for X in either equation. I'm starting to develop a bad habit of doing all the work for the questions I help on, so I will leave the computations to you unless you are still stuck.
Circles are congruent, that is same radius, exactly equal. And furthermore, QR and LN are the same length, as are WV and OP. So that means: x + y = 6 10 = 3x - y
then what do i do?
solve the first equation for x x = 6 - y plug into the second equation 10 = 3(6 - y) - y can you solve that for y?
x = 6 - y 10 = 3(6 - y) – y 10=18-3y-y ?
We have 10 = 3(6-y) - y So I see you multiplied it out. Yes, that's the first step.
Now what?
subtract 10?
why not put your y's together
3y+y?
or +3y?
We have -3y -y We can add those -4y on the right side, or put 4y on the left side, your choice.
4y + 10 = 18 Now subtract 10.
x = 6 - y 10 = 3(6 - y) – y 10=18-3y-y 4y + 10 = 18 -10 -10 4y=8 4 4 y=4
would it look like this?
Good. 4y = 8 y = 4 ??? Doesn't sound right.
Basic arithmetic: 4 times 2 = 8. Right? So y = 2.
thanks!!! :)
Still need x. And remember to check your work. yw
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