Answer:
A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S.
The given equation consisting of variable , m is
![\frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})](https://tex.z-dn.net/?f=%5Cfrac%7B2%20m%7D%7B2%20m%2B3%7D%20-%5Cfrac%7B2%20m%7D%7B2%20m-3%7D%3D1%5C%5C%5C%5C%202%20m%5B%5Cfrac%7B1%7D%7B2%20m%2B3%7D%20-%5Cfrac%7B1%7D%7B2%20m-3%7D%5D%3D1%5C%5C%5C%5C%202%20m%5Ctimes%20%5Cfrac%7B%5B2%20m-3%20-2%20m-%203%5D%7D%7B4m%5E2-9%7D%3D1%5C%5C%5C%5C%20-6%20%5Ctimes%202%20m%3D4%20m%5E2%20-9%5C%5C%5C%5C%204%20m%5E2%20%2B1%202%20m%20-9%3D0%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%5Csqrt%7B12%5E2-4%20%5Ctimes%204%20%5Ctimes%20%28-9%29%7D%7D%7B2%5Ctimes%204%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%20%5Csqrt%20%7B144%2B144%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%20%5Csqrt%20%7B288%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%2012%20%5Csqrt%7B2%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B3%7D%7B2%7D%5Ctimes%28-1%20%5Cpm%20%5Csqrt%7B2%7D%29)
None of the two solution
, is extraneous.
Here, L.H.S= R.H.S
Option A: 0→ extraneous
Answer:
179.3
Step-by-step explanation:
hope this helps
have a good day
(-8, -11) because the points are just the opposite sign when reflecting over the x-axis
<h2>
Answer: (5, 2)</h2>
<h2>
System of equations:</h2>
2x - y = 8
3x + y = 17
<h2>Solve for x:</h2>
2x - y = 8
3x + y = 17
________ +
5x = 25
x = 5
<h2>Plug in x to one of the equations to solve for Y</h2>
2(5) - y = 8
10 - y = 8
-y = -2
y = 2
You subtract 5 to the other side so it'll read -6n=-4 then divide -6 to the other side and it'll be 2/3
