![ax + by = a + b \\ ax + b(x + y) = a + b ax + by = a + b \\ ax + b(x + y) = a + b](https://tex.z-dn.net/?f=ax+%2B+by+%3D+a+%2B+b++%5C%5C+ax+%2B+b%28x+%2B+y%29+%3D+a+%2B+b)
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Answered by
3
Let us consider,
ax+by=a+b---(1)
ax+b(x+y)=a+b----(2)
Equate(1) & (2)..,,
ax+by=ax+b(x+y)
by=b(x+y)
y=b(x+y) / b
y=x+y
x=y-y
x=0
substitute x=0 in (1)..,
ax+by=a+b
a(0)+by=a+b
by=a+b
y=a+b/b
ax+by=a+b---(1)
ax+b(x+y)=a+b----(2)
Equate(1) & (2)..,,
ax+by=ax+b(x+y)
by=b(x+y)
y=b(x+y) / b
y=x+y
x=y-y
x=0
substitute x=0 in (1)..,
ax+by=a+b
a(0)+by=a+b
by=a+b
y=a+b/b
kjha76776:
thanks but what is value of y
Answered by
1
ax+by=a+b....(i)
ax+b(x+y)=a+b.....(ii)
In (i),
![ax = a + b - by \\ substituting \: value \: in \: ii \\ a + b - by + bx + by = a + b \\ = > bx = 0 \\ = > x = 0 \\ ax = a + b - by \\ substituting \: value \: in \: ii \\ a + b - by + bx + by = a + b \\ = > bx = 0 \\ = > x = 0 \\](https://tex.z-dn.net/?f=ax+%3D+a+%2B+b+-+by+%5C%5C+substituting+%5C%3A+value+%5C%3A+in+%5C%3A+ii+%5C%5C+a+%2B+b+-+by+%2B+bx+%2B+by+%3D+a+%2B+b+%5C%5C+%3D+%26gt%3B+bx+%3D+0+%5C%5C+%3D+%26gt%3B+x+%3D+0+%5C%5C+)
ax+b(x+y)=a+b.....(ii)
In (i),
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