How many solutions does this linear system have?

Mathematics

1 answer:

gayaneshka [121]3 years ago

4 0

Answer:

One solution: (2.5,0)

Step-by-step explanation:

We can use substitution for the solution. Substitute y=2x-5 to -8x-4y=-20 so that you end up with -8x-8x+20=-20. Next you want to add like terms which will be -16x+20=-20, next you want isolate x by subtracting 20 from both sides and leaves you with -16x=-40. Divide -16 on both sides to fully isolate x and will leave you with x=2.5. Now substiture in 2.5 for x in y=2x-5 to get y=2(2.5)-5 which will then lead to y=0.

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Shkiper50 [21]

Answer:

$x=1$

Step-by-step explanation:

$~~~~~x = \dfrac{1}{2 - \dfrac 1{2-\dfrac{1}{2-x}}}\\\\\\\implies x = \dfrac{1}{2 - \dfrac 1{\dfrac{4-2x-1}{2-x}}}\\\\\\\implies x= \dfrac{1}{2 - \dfrac{2-x}{3-2x}}\\\\\\\implies x = \dfrac{1}{\dfrac{6-4x-2+x}{3-2x}}\\\\\\\implies x = \dfrac{3-2x}{4-3x}\\\\\\\implies 4x -3x^2 = 3-2x\\\\\\\implies 3x^2 -4x +3-2x = 0\\\\\\\implies 3x^2 -6x +3 = 0\\\\\\\implies 3(x^2 -2x +1) =0\\\\\\\implies x^2 -2x +1 = 0\\\\\\\implies (x-1)^2 = 0\\\\\\\implies x -1 = 0\\\\\\\implies x = 1$