Question

Solve the following pair of equations for x and y:
 $\dfrac{10}{x+y}+\dfrac{2}{x-y}=4$ and $\dfrac{15}{x+y}-\dfrac{5}{x-y}=-2$ .

Answer 1

Hint:
1) Substitute $\dfrac{1}{x+y}=a$ and $\dfrac{1}{x-y}=b$ , to get two linear equations in variables a and b.
2) A pair of simultaneous linear equations in two variables can be solved by either Cramer's rule, by elimination, or by substitution.
3) After finding the values of a and b, we need to solve the simultaneous linear equations $x+y=\dfrac{1}{a}$ and $x-y=\dfrac{1}{b}$ .

Complete step by step solution:
Let's substitute $\dfrac{1}{x+y}=a$ and $\dfrac{1}{x-y}=b$ , to get two linear equations:
10a + 2b = 4 ... (1)
15a - 5b = -2 ... (2)
Let's use the method of elimination to solve these equations. In order to eliminate one of the variables, we will make their coefficients equal by multiplying the equations and then add / subtract the equations.
On multiplying equation (1) by 3 and equation (2) by 2, to make the coefficient of a 30 in both of them, we will get:
30a + 6b = 12 ... (3)
30a - 10b = -4 ... (4)
Subtracting equation (4) from equation (3), we get:
(30a + 6b) - (30a - 10b) = (12) - (-4)
⇒ 16b = 16
⇒ b = 14
Putting this value of b in any of the above equations, let's say equation (1), we get:
10a + 2(1) = 4
⇒ 10a = 2
⇒ $a=\dfrac{1}{5}$
Now, using these values of a and b in $\dfrac{1}{x+y}=a$ and $\dfrac{1}{x-y}=b$ , we get the following set of equations:
 $\dfrac{1}{x+y}=\dfrac{1}{5}$
⇒ x + y = 5 ... (5)
 $\dfrac{1}{x-y}=1$
⇒ x - y = 1 ... (6)
Since, the coefficients of y have the same magnitude and opposite signs, we can add the equations (5) and (6) to eliminate y:
(x + y) + (x - y) = (5) + (1)
⇒ 2x = 6
⇒ x = 3
Putting this value of x in, say, equation (5), we get:
3 + y = 5
⇒ y = 2

Hence, the value of x is 3 and of y is 2.

Note:
The solutions to the equations ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$ can be determined using Cramer's rule:
 $D=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} \\ {{b}_{1}} & {{b}_{2}} \\ \end{matrix} \right|$ , ${{D}_{x}}=\left| \begin{matrix} {{c}_{1}} & {{c}_{2}} \\ {{b}_{1}} & {{b}_{2}} \\ \end{matrix} \right|$ , ${{D}_{y}}=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} \\ {{c}_{1}} & {{c}_{2}} \\ \end{matrix} \right|$ .
The values of the variables x and y are given by: $x=\dfrac{{{D}_{x}}}{D}$ and $y=\dfrac{{{D}_{y}}}{D}$ .
If, D = 0, the system is:
Either consistent and has infinitely many solutions. In this case ${{D}_{x}}={{D}_{y}}=0$ .
Or it is inconsistent, i.e. it has no solutions. In this case ${{D}_{x}}\ne 0$ and ${{D}_{y}}\ne 0$