youngnephew34
  • youngnephew34
What is the value of y in rhombus LMNO A)4 B)6 C)12 D)24
Mathematics
jamiebookeater
  • jamiebookeater
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youngnephew34
  • youngnephew34
phi
  • phi
try to write an equation that shows the two halves of a diagonal are equal
youngnephew34
  • youngnephew34
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youngnephew34
  • youngnephew34
hold on
youngnephew34
  • youngnephew34
Y+12=6x
phi
  • phi
yes, now write the other equation (for the other diagonal)
youngnephew34
  • youngnephew34
\[Y=3x\]
phi
  • phi
we have two equations and two unknowns (x and y are the unknowns) do you know how to solve them? Y+12=6x y= 3x
phi
  • phi
one way is to use "substitution", the other way is "elimination"
youngnephew34
  • youngnephew34
For the Y+12=6x don't you suppose to subtract one side from another
phi
  • phi
in this problem, substitution works pretty well. start with the equations: y+12=6x y= 3x notice the 2nd equation says "y is the *same thing* as 3x" we use that to replace the y in the first equation with 3x (erase the y and put in 3x instead) can you do that?
youngnephew34
  • youngnephew34
you will substitute the 3x in to y right
phi
  • phi
what equation do you get ?
phi
  • phi
erase the y and put in 3x instead
youngnephew34
  • youngnephew34
so 3x+12=6x
phi
  • phi
yes. now we have just 1 unknown... that is good. to solve I would start by adding -3x to both sides, like this 3x - 3x + 12 = 6x - 3x now simplify
phi
  • phi
on the right side you have 6 x's take away 3 x's. how many x's do we have?
youngnephew34
  • youngnephew34
3x
phi
  • phi
ok, but write out the whole equation (so we don't get confused) 3x - 3x + 12 = 3x on the right side, we have 3x's take away 3 x's
phi
  • phi
*left side
youngnephew34
  • youngnephew34
3x+12=3x
phi
  • phi
almost. 3x - 3x is not 3x
youngnephew34
  • youngnephew34
so really 12=3x
phi
  • phi
yes last step. divide both sides by 3
youngnephew34
  • youngnephew34
x=4
phi
  • phi
the question asks for "y" not x but from the equations we know y= 3x can you find y (knowing x is 4) ?
youngnephew34
  • youngnephew34
Y=4 srry bout that
phi
  • phi
no, you were correct the first time x=4 but we can find y using y= 3*x
youngnephew34
  • youngnephew34
so divide 4=3x
phi
  • phi
y= 3x says y is the same thing as 3 times x if we knew what x is (as a number) we would multiply it by 3 and we would get y (as a number) but from the earlier steps you found x is 4 that means we can find y using that equation. can you try?
phi
  • phi
in other words, in y= 3*x erase the x and put in 4 instead (because x is the same thing as 4)
youngnephew34
  • youngnephew34
Instead you would do 4x3 right
phi
  • phi
yes, but write it 4*3 (because using x for multiply is confusing... it might mean "x" the variable)
youngnephew34
  • youngnephew34
ok so 4*3=12
phi
  • phi
right, y=3x y=3*4 y=12 this problem took lots of steps. See if you can go over the steps, and try to understand the way we found y is 12.
youngnephew34
  • youngnephew34
so 3x+12+6x 3x-6x=3x 12/3=3/3 12/3=4 x=4 4x3=12
phi
  • phi
3x+12=6x then add -3x to both sides 3x-3x + 12 = 6x-3x simplify to get 0+12 = 3x or 12=3x <-- notice we have 3x, not just 3) 4=x
phi
  • phi
then use y=3x when x is 4 then y = 3*4 =12
youngnephew34
  • youngnephew34
ok got it and one more to check

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