How to Solve bu using elimination method ?

Question

How to Solve bu using elimination method ?

$\frac{5}{x + y} + \frac{2}{x - y} = 3$ $5/(x+y) + 2/(x-y) = 3$
$\frac{20}{x + y} - \frac{3}{x - y} = 1$ $20/(x+y) - 3/(x-y) = 1$

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Answer 1 · 2018-06-04T16:29:44

$x = - \frac{11}{2}, y = - \frac{9}{2}$ $x=-11/2,y=-9/2$

Explanation:

Substituting
$a = \frac{1}{x + y}, b = \frac{1}{x - y}$ $a=1/(x+y),b=1/(x-y)$
so we have the System
$5 a + 2 b = 3$ $5a+2b=3$
$20 a - 3 b = 1$ $20a-3b=1$
Multiplying the first equation by $- 4$ $-4$ and adding to the second we get
$b = - 1$ $b=-1$
so
$a = - \frac{1}{10}$ $a=-1/10$
Then we have
$1 = - x + y$ $1=-x+y$
$- 10 = x + y$ $-10=x+y$
from here we get
$x = - \frac{11}{2}$ $x=-11/2$
$y = - \frac{9}{2}$ $y=-9/2$

Answer 2 · 2018-06-04T17:09:11

$x = 3 and y = 2$ $x = 3 and y = 2$

Explanation:

$\frac{5}{x + y} + \frac{2}{x - y} = 3$ $5/(x + y) + 2/(x - y)=3$

$\frac{20}{x + y} - \frac{3}{x - y} = 1$ $20/(x + y)-3/(x - y)= 1$

Let;

$\frac{1}{x + y} = a and \frac{1}{x - y} = b$ $1/(x + y) = a and 1/(x - y) = b$

Therefore;

$\frac{5}{x + y} + \frac{2}{x - y} = 3$ $5/(x + y) + 2/(x - y)=3$

$5 (\frac{1}{x + y}) + 2 (\frac{1}{x - y}) = 3$ $5(1/(x + y)) + 2(1/(x - y))=3$

$5 a + 2 b = 3 - - - e q n 1$ $5a + 2b = 3 - - - eqn1$

Similarly..

$\frac{20}{x + y} - \frac{3}{x - y} = 1$ $20/(x + y)-3/(x - y)= 1$

$20 (\frac{1}{x + y}) - 3 (\frac{1}{x - y}) = 1$ $20(1/(x + y))-3(1/(x - y)) = 1$

$20 a - 3 b = 1 - - - e q n 2$ $20a - 3b = 1 - - - eqn2$

Using Elimination Method!

$5 a + 2 b = 3 - - - e q n 1$ $5a + 2b = 3 - - - eqn1$

$20 a - 3 b = 1 - - - e q n 2$ $20a - 3b = 1 - - - eqn2$

Multiply $e q n 1$ $eqn1$ by $3$ $3$ and $e q n 2$ $eqn2$ by $2$ $2$

$3 (5 a + 2 b = 3)$ $3(5a + 2b = 3)$

$2 (20 a - 3 b = 1)$ $2(20a - 3b = 1)$

$15 a + 6 b = 9 - - - e q n 3$ $15a + 6b = 9 - - - eqn3$

$40 a - 6 b = 2 - - - e q n 4$ $40a - 6b = 2- - - eqn4$

Adding both $e q n 3 and e q n 4$ $eqn3 and eqn4$ together..

$(15 a + 40 a) + (6 b + (- 6 b)) = 9 + 2$ $(15a + 40a) + (6b + (-6b))= 9 + 2$

$55 a + 6 b - 6 b = 11$ $55a + 6b - 6b = 11$

$55 a = 11$ $55a = 11$

$a = \frac{11}{55}$ $a = 11/55$

$a = \frac{1}{5}$ $a = 1/5$

Substituting the value of $a$ $a$ into $e q n 1$ $eqn1$

$5 a + 2 b = 3 - - - e q n 1$ $5a + 2b = 3 - - - eqn1$

$5 (\frac{1}{5}) + 2 b = 3$ $5(1/5) + 2b = 3$

$cancel5(1/cancel5) + 2b = 3$

$1 + 2b= 3$

$2b = 3 - 1$

$2b = 2$

$b = 2/2$

$b = 1$

But;

$1/(x + y) = a and 1/(x - y) = b$

$a = 1/(x + y)$

$1/5 = 1/(x + y)$

Cross multiplying;

$1(x + y) = 1(5)$

$x + y = 5 - - - eqn5$

Similarly..

$b = 1/(x - y)$

$1 = 1/(x- y)$

$1/1 = 1/(x- y)$

Cross multiplying;

$1(x - y) = 1(1)$

$x - y = 1 - - - eqn6$

Solving simultaenously again..

$x + y = 5 - - - eqn5$

$x - y = 1 - - - eqn6$

Using Elimination Method!

Adding $eqn5 and eqn6$ together;

$(x + x) + (y +(-y)) = 5 + 1$

$2x + y - y = 6$

$2x = 6$

$x = 6/2$

$x = 3$

Substituting the value of $x$ into $eqn6$

$x + y = 5 - - - eqn5$

$3 + y = 5$

Collecting like terms;

$y = 5 - 3$

$y = 2$

Hence;

$x = 3 and y = 2$

Answer 3 · 2018-06-04T19:38:32

$x=3,y=2$

Explanation:

$5/(x+y)+2/(x-y)=3$
$20/(x+y)-3/(x-y)=1$

Let x+y be a and x-y be b

$:.5/a+2/b=3$ ----------(1)

$:.20/a-3/b=1-----------(2)$

$:.(1) xx 4$

$:.20/a+8/b=12$ ------(3)

$:.(2)-(3)$

$:.-11/b=-11/1$

$:.b/-11=1/-11$

multiply both sides by-11

$:.b=-11/-11$

$:.b=1$

substitute b=1 in (1)

$:.5/a+2=3$

$:.5/a=3-2$

$:.5/a=1$

$:.a/5=1$

multiply both sides by 5

$:.a=5$

$:.a=x+y=5$ -----(4)

$:.b=x-y=1$ ------(5)

$:.(4)+(5)$

$:.2x=6$

$:.x=6/2$

$:.x=3$

$:.3+y=5$

$:.y=5-3$ in (4)

$:.y=2$

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How to Solve bu using elimination method ?

$\frac{5}{x + y} + \frac{2}{x - y} = 3$ $5/(x+y) + 2/(x-y) = 3$
$\frac{20}{x + y} - \frac{3}{x - y} = 1$ $20/(x+y) - 3/(x-y) = 1$

3 Answers

Explanation:

Explanation:

Explanation:

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How to Solve bu using elimination method ?

5/(x+y) + 2/(x-y) = 35x+y+2x−y=3 5/(x+y) + 2/(x-y) = 3 20/(x+y) - 3/(x-y) = 120x+y−3x−y=1 20/(x+y) - 3/(x-y) = 1

3 Answers

Explanation:

Explanation:

Explanation:

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Impact of this question

$\frac{5}{x + y} + \frac{2}{x - y} = 3$ $5/(x+y) + 2/(x-y) = 3$
$\frac{20}{x + y} - \frac{3}{x - y} = 1$ $20/(x+y) - 3/(x-y) = 1$