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3 years ago

(x+y,x-2y)=(4,1), find x and y

Mathematics

2 answers:

dimulka [17.4K]3 years ago

6 0

Answer:

x = 3, y = 1

Step-by-step explanation:

Equate the x and y coordinates of the given points

x + y = 4 → (1)

x - 2y = 1 → (2)

Subtract (2) from (1) term by term to eliminate x

0 + y - (- 2y) = 4 - 1

y + 2y = 3

3y = 3 ( divide both sides by 3 )

y = 1

Substitute y = 1 into (1) for corresponding value of x

x + 1 = 4 ( subtract 1 from both sides )

x = 3

gtnhenbr [62]3 years ago

5 0

Answer:

x =3 and y =1

Step-by-step explanation:

( x+y , x-2y) = (4 , 1)

compare their corresponding elements

x+y = 4

y = 4 - x equation (i)

x-2y =1

substitute the value of x

x - 2(4 - x) =1

x - 8 + 2x =1

3x = 1 +8

x = 9/3

x = 3

Now substitute the value of x in equation (i)

y = 4 - x

y = 4 - 3

y = 1

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4 years ago

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Vadim26 [7]

The value of x is $x=\frac{-1+\sqrt{37}}{12}$ and $x=\frac{-1-\sqrt{37}}{12}$

Step-by-step explanation:

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$\frac{1}{x}-\frac{2}{3}-4 x=0$

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Using quadratic formula, we can solve for x.

$\begin{aligned}x &=\frac{-2 \pm \sqrt{2^{2}-4 \cdot 12 \cdot(-3)}}{2 \cdot 12} \\&=\frac{-2 \pm \sqrt{4+144}}{2 \cdot 144} \\&=\frac{-2 \pm \sqrt{148}}{24} \\&=\frac{-2 \pm 2 \sqrt{37}}{24}\end{aligned}$

Taking out common term 2, we get,

$\begin{array}{l}{x=\frac{-2(1 \pm \sqrt{37})}{24}} \\{x=\frac{-1 \pm \sqrt{37}}{12}}\end{array}$

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3 years ago