A) 3,2
B) 2,3
C) 6,3
D) 3,4
Correct Answer: A
Solution :
Given equations are |
\[x+8y=19\] ...(i) |
and \[2x+11y=28\] ...(ii) |
From Eq.(i), \[x=19-8y\] ...(iii) |
On substituting \[x=19-8y\]in Eq. (ii), we get |
\[2\left( 19-8y \right)+11y=28\] |
\[\Rightarrow \,\,38-16y+11y=28\] |
\[\Rightarrow \,\,\,5y=38-28=10\] |
\[\Rightarrow \,\,\,\,y=\frac{10}{5}=2\] |
Now, on putting y = 2 in Eq. (iii), we get |
\[x=19-8\times 2\] |
\[\Rightarrow \,\,\,x=19-16=3\] |
Thus, \[x=3\]and y = 2 is the required solution. |
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