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Neporo4naja [7]
3 years ago
11

X-2y=-25 3x-y=0 need help solving

Mathematics
1 answer:
lukranit [14]3 years ago
8 0
x-2y=-25\\
3x-y=0\\\\
x-2y=-25\\
y=3x\\\\
x-2\cdot3x=-25\\
x-6x=-25\\
-5x=-25\\
x=5\\\\
y=3\cdot5\\
y=15

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the equation is -2x² = 4-3 (x + 1) and the question is justify that it is a 2nd degree equation with the unknown x complete.
Serggg [28]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the equation

-2x² = 4-3 (x + 1)

-2x² = 4-3x-3

-2x² = -3x -7

0 = 2x² -3x -7

We know that the degree of the equation is the highest power of x variable in the given equation.

In the equation 0 = 2x² -3x -7 the highest power of x variable in the given equation is 2.

Thus, the degree of the equation is 2.

Also in the equation 0 = 2x² -3x -7, the unknown variable is 'x'.

Let us determine the value 'x'

2x² -3x -7 = 0

Add 7 to both sides

2x^2-3x-7+7=0+7

2x^2-3x=7

Divide both sides by 2

\frac{2x^2-3x}{2}=\frac{7}{2}

x^2-\frac{3x}{2}=\frac{7}{2}

Add (-3/4)² to both sides

x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{7}{2}+\left(-\frac{3}{4}\right)^2

x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{65}{16}

\left(x-\frac{3}{4}\right)^2=\frac{65}{16}

\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

solving

x-\frac{3}{4}=\sqrt{\frac{65}{16}}

x-\frac{3}{4}=\frac{\sqrt{65}}{\sqrt{16}}

x-\frac{3}{4}=\frac{\sqrt{65}}{4}

Add 3/4 to both sides

x-\frac{3}{4}+\frac{3}{4}=\frac{\sqrt{65}}{4}+\frac{3}{4}

x=\frac{\sqrt{65}+3}{4}

similarly solving

x-\frac{3}{4}=-\sqrt{\frac{65}{16}}

x=\frac{-\sqrt{65}+3}{4}

So the solution of the equation will have the values of x such as:

x=\frac{\sqrt{65}+3}{4},\:x=\frac{-\sqrt{65}+3}{4}

6 0
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crimeas [40]

Answer:

if each tube of paints costs a dollar he will get 25 because of the 15 dollars added to the 10 dollars

Step-by-step explanation:

hope this helps

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

please give me brainlest to me please

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