Question

Simultaneous Linear Equations​

Answer 1

Answer

• The value of x = 4 and y = 3.

Given

• $\tt \cfrac{14}{x + y} + \cfrac{3}{x - y} = 5$

• $\tt \cfrac{21}{x + y} - \cfrac{1}{x - y} = 2$

To Find

• The value of x and y.

Step By Step Explanation

Assumption :

Let us assume that $\sf\cfrac{1}{x+y}$ be u and $\sf\cfrac{1}{x-y}$ be v.

Equation :

Then the equation will be ⤵

$\sf14u + 3v = 5$ and $\sf21u - v = 2$

Solution of equation :

Let's solve the above equation.

$\sf \: \: \: \: \: \: \: 14u + 3v = 5 \\ \sf3 \times (21u - v = 2) \downarrow \\ \\ \tt14u + 3v = 5 \\ \tt 63u - 3v = 6 \\ - - - - - - \\ \tt 77u = 11 \implies \: u \: = \cfrac{ \cancel{11}}{ \cancel{77}} \\ \\ \implies \bold{ \green{u = \cfrac{1}{7}}} \\ \\ \tt \cancel{14} \times \cfrac{1}{ \cancel7} + 3v = 5 \implies 2 + 3v = 5 \\ \\ \implies \bold{ \pink{v = 1}}\\ \\ \bold{u = \cfrac{1}{7} \: \: \: and \: \: \: v =1 }$

By substituting the value :

Let's substituting the values of u and v. To find the value of x and y.

So let's do it !!

$\longmapsto \sf \cfrac{1}{x + y} = u \\ \\ \longmapsto \sf\cfrac{1}{x + y} = \cfrac{1}{7} \\ \\ \longmapsto \bold{ x + y = 7}\: \: \: \: \: \: \: \: = > eq. 1 \\ \\ \\ \longmapsto \sf \cfrac{1}{x - y} = v \\ \\ \longmapsto \sf\cfrac{1}{x - y} = 1 \\ \\\longmapsto \bold{ x - y = 1} \: \: \: \: \: \: \: = > eq.2$

Now, x and y will be ⤵

$\sf{x + y = 7} \\ \sf{ x - y = 1} \\ - - - - - \\ \sf{ 2x = 8} \implies \: { \boxed{\bold{ \red{x = 4}}}} \: \: \: \: \: \bigstar\\ \\ \\ \sf{x + y = 7 \implies4 + y = 7} \\ \\ \sf{ y = 7 - 4} \implies \: { \boxed{\bold{ \purple{ y = 3}}}} \: \: \: \: \: \bigstar$

Therefore, the value of x = 4 and y = 3.

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