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Byju's Answer
Standard IX
Mathematics
Addition and Subtraction of Algebraic Equations
vii 2 × 15+y ...
Question
(vii)
2
x
15
+
y
5
=
4
;
x
3
=
y
2
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Solution
2
x
15
+
y
5
=
4
⇒
2
x
+
3
y
=
60
.
.
.
(
i
)
x
3
=
y
2
⇒
2
x
-
3
y
=
0
.
.
.
(
ii
)
Adding
eq
(
i
)
and
(
ii
)
,
we get
:
2
x
+
3
y
=
60
2
x
-
3
y
=
0
4
x
=
60
∴
x
=
15
Substituting
x
=
15
in
eq
(
ii
)
,
we get:
2
×
15
-
3
y
=
0
⇒
30
-
3
y
=
0
∴
y
=
10
Hence
,
x
=
15
and
y
=
10
is
the
solution
of
the
given
equations
.
Suggest Corrections
0
Similar questions
Q.
Simplify
[
(
x
y
)
2
]
2
×
x
3
y
3
x
4
y
5
Q.
Simplify:
x
5
y
3
−
x
3
y
5
x
3
y
3
Q.
Using binomial theorem, expand
{
(
x
+
y
)
5
+
(
x
−
y
)
5
}
and hence find the value of
{
(
√
2
+
1
)
5
+
(
√
2
−
1
)
5
}
.
Q.
Eliminate
x
,
y
,
z
from the equations
x
+
y
+
z
=
0
,
x
2
+
y
2
+
z
2
=
a
2
,
x
3
+
y
3
+
z
3
=
b
3
,
x
5
+
y
5
+
z
5
=
c
5
.
Q.
Write down and simplify:
The
5
t
h
term of
⎛
⎝
x
3
2
a
1
2
−
y
5
2
b
3
2
⎞
⎠
8
.
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