solve by elemination method. (1) 5/x+y - 2/ x-y = -1, 15/x+y + 7/ x-y =10
Answers
Answer:
Given:
\begin{gathered}\frac { 5 }{ x+y } -\quad \frac { 2 }{ x-y } = -1\\ \frac { 15 }{ x+y } +\frac { 7 }{ x-y } \quad =10.\end{gathered}
x+y
5
−
x−y
2
=−1
x+y
15
+
x−y
7
=10.
Solution:
Concept: “If the number of equations given and the number of variables in the system of equations are equal, then there is an exact or unique solution exists.”
Let \frac{1}{(x+y)} = a \quad and\quad\frac{1}{(x-y)} = b;
(x+y)
1
=aand
(x−y)
1
=b;
On substitution, the system of equations become
5a - 2b = -1 ---------------(1)
15a +7b = 10. ---------------(2)
To eliminate a, multiply the (1) equation by factor 3, which gives:
15a - 6b = -3 ---------------(3)
Elimination method (Eliminating one variable and finding value of other variable).
(2) – (3) gives:
(15a + 7b) - (15a – 6b) = 10-(-3)
15a + 7b – 15a + 6b = 10 + 3
13b = 13
b = 1
Substitute the value of b in any of (1) or (3), which gives:
a = \frac{1}{5}.a=
5
1
.
To find x and y:
From the assumption made, substitute a and b values:
\frac{1}{(x+y)} = aand\frac{1}{(x-y)} = b.
(x+y)
1
=aand
(x−y)
1
=b.
\frac{1}{(x+y)} = \frac{1}{5};
(x+y)
1
=
5
1
;
\frac{1}{(x-y)} =\frac{1}{1};
(x−y)
1
=
1
1
;
x + y = 5; x – y = 1
By adding the above equations:
(x + y) + (x-y) = 5+1
2x = 6
x = 3.
On substitution,
3 + y = 5
y = 2
values are x = 3 and y = 2;"