Solve the following systems of equations:
99x + 101y = 499
101x + 99y = 501
Answers
Concept :
When the coefficients of x and y in one equation interchanged in the other:
Step 1. Add the given equations and from it obtain an equation of the form x+ y = a.
Step 2. Subtract the given equations and from it obtain an equation of the form x -y = b.
Now , x + y = a and x - y = b can be solved easily.
SOLUTION:
99x + 101y = 499 …………...( 1 )
101x + 99y = 501 …………...( 2 )
On Adding Equation 1 and 2 :
99x + 101y = 499
101x + 99y = 501
---------------------------
200x + 200y = 1000
--------------------------
200 (x + y) = 1000
x + y = 1000/200
x + y = 5 ……………….(3)
On Subtracting equation 1 and 2 :
99x + 101y = 499
101x + 99y = 501
(-) (-) (-)
--------------------------
-2x + 2y = - 2
-----------------------------
-2(x - y) = - 2
x - y = - 2/-2
x - y = 1…………...( 4 )
On adding equation 3 and 4
x + y = 5
x - y = 1 [By elimination method]
--------------
2x = 6
x = 6/2
x = 3
On substituting x = 3 in eq 3 :
x + y = 5
3 + y = 5
y = 5 - 3
y = 2
Hence ,the solution of the given system of equation is x = 3 and y = 2.
Hope this answer will help you…
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Answer:
--------------------------
200 (x + y) = 1000
x + y = 1000/200
x + y = 5 ……………….(3)
On Subtracting equation 1 and 2 :
99x + 101y = 499
101x + 99y = 501
(-) (-) (-)
--------------------------
-2x + 2y = - 2
-----------------------------
-2(x - y) = - 2
x - y = - 2/-2
x - y = 1…………...( 4 )
On adding equation 3 and 4
x + y = 5
x - y = 1 [By elimination method]
--------------
2x = 6
x = 6/2
x = 3
On substituting x = 3 in eq 3 :
x + y = 5
3 + y = 5
y = 5 - 3
y = 2