Answer:
(5, - 4 )
Step-by-step explanation:
Given the 2 equations
2x + 3y = - 2 → (1)
3x - y = 19 → (2)
Multiplying (2) by 3 and adding to (1) will eliminate the y- term
9x - 3y = 57 → (3)
Add (1) and (3) term by term to eliminate y
(2x + 9x) + (3y - 3y) = (- 2 + 57), that is
11x = 55 ( divide both sides by 5 )
x = 5
Substitute x = 5 into either of the 2 equations and solve for y
Substituting into (1)
2(5) + 3y = - 2
10 + 3y = - 2 ( subtract 10 from both sides )
3y = - 12 ( divide both sides by 3 )
y = - 4
Solution is (5, - 4 )
The value of GK = 18.6, GM = 14.5 AND JZ = 19.9.
From the figure:
MZ is a perpendicular bisector of ∆GHJ.
GM = MJ
GM = 14.5
KZ is a perpendicular bisector of ∆GHJ.
GK = KH
GK = 18.6
Z is the circumcenter of ∆GHJ.
JZ = GZ
JZ = 19.9
Therefore the value of GK = 18.6, GM = 14.5 AND JZ = 19.9.
Learn more about the perpendicular bisector here:
brainly.com/question/24753075
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Answer:
The given expression is equivalent to -x
That is ![\frac{1}{5}(5x-20)-\frac{1}{2}(4x-8)=-x](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D%285x-20%29-%5Cfrac%7B1%7D%7B2%7D%284x-8%29%3D-x)
Now Verified that the two expressions are equivalent
Step-by-step explanation:
Given that Genevieve wants to verify that One-fifth (5 x minus 20) minus one-half (4 x minus 8) is equivalent to Negative x
It can be written as below :
![\frac{1}{5}(5x-20)-\frac{1}{2}(4x-8)=-x](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D%285x-20%29-%5Cfrac%7B1%7D%7B2%7D%284x-8%29%3D-x)
Now we can verify the expression is equivalent or not :
Taking LHS
![\frac{1}{5}(5x-20)-\frac{1}{2}(4x-8)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D%285x-20%29-%5Cfrac%7B1%7D%7B2%7D%284x-8%29)
( by using the distributive property )
![=x-4-2x+4](https://tex.z-dn.net/?f=%3Dx-4-2x%2B4)
( adding the like terms )
=RHS
Therefore LHS=RHS
Therefore the given expression is equivalent to -x
That is ![\frac{1}{5}(5x-20)-\frac{1}{2}(4x-8)=-x](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D%285x-20%29-%5Cfrac%7B1%7D%7B2%7D%284x-8%29%3D-x)
Now Verified that the two expressions are equivalent