Solve the following systems of equations:
21x + 47y = 110
47x + 21y = 162
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:
21x + 47y = 110 …………...( 1 )
47x + 21y = 162…………...( 2 )
On Adding Equation 1 and 2 :
21x + 47y = 110
47x + 21y = 162
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68x + 68y = 272
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68 (x + y) = 272
x + y = 272/68
x + y = 4 ……………….(3)
On Subtracting equation 1 and 2 :
21x + 47y = 110
47x + 21y = 162
(-) (-) (-)
--------------------------
-26x + 26y = - 52
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-26(x - y) = - 52
x - y = - 52/-26
x - y = 2…………...( 4 )
On adding equation 3 and 4 :
x + y = 4
x - y = 2 [By elimination method]
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2x = 6
x = 6/2
x = 3
On substituting x = 3 in eq 3 :
x + y = 4
3 + y = 4
y = 4 - 3
y = 1
Hence ,the solution of the given system of equation is x = 3 and y = 1.
Hope this answer will help you…
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Step-by-step explanation:
We have
21x + 47y = 110 (1)
47x + 21y = 162 (2)
Adding Equations (1) and (2), we have
68x + 68y = 272
or x + y = 4 (5)
Subtracting Equation (1) from Equation (2), we have
26x – 26y = 52
or x – y = 2 (6)
On adding and subtracting Equations (5) and (6), we get
x = 3, y = 1