Step-by-step explanation:
e method of solving this type of equations by reducing to linear formmy variable substitution is used.
Observe the key pattern in the two equations:
1
x
and
1
y
appear in both equations and these are independent of each other.
As if these two expressions behave like two independent variables.
At this point you see the possibility—why not replace each of these two expressions by a distinct variable!
You decide to substitute dummy variables (to be used temporarily),
p=
1
x
, and
q=
1
y
.
And the two equations are immediately transformed by these substitutions to,
2p−3q=
4
7
and,
5p+7q=
7
2
.
Now you have two equations in p and q that are linear and you know more than one method to solve such a pair of linear equations.
If a unique solution to this pair of equations exists, by solving the equations you would get the values of p and q and hence of x and y.
Excluding graphical method, you may use any of the other methods to solve the pair of linear equations in p and q.
As we consider the method of elimination by fraction multiplier the quickest with least number of steps, we'll adopt this method.
Identifying the simplest term 2p to eliminate, multiply the first equation by
5
2
. Result is,
5p−
15
2
q=
10
7
.
Subtract this result from the second equation to eliminate p. Result is,
29
2
q=
7
2
−
10
7
=
29
14
,
Or, q=
1
7
=
1
y
.
So, y=7.
Substitute the value of q in first equation, 2p−3q=
4
7
to get,
2p−
3
7
=
4
7
,
Or, 2p=1,
Or, p=
1
2
=
1
x
.
So, x=2.
Answer: x=2, y=7.
Verification: Substitute these two values in the given equations. Results are,
1−
3
7
=
4
7
, and
5
2
+1=
7
2
.
Solution verified.
Reducing method Class 10: Conditions under which a pair of non-linear equations can be reduced to linear form
Let us specify the conditions under which you can apply this novel method of solving two non-linear equations by expression substitution with dummy variables.
To solve two non-linear equations in two variables by reducing to linear forms, you must be able to present the equations in terms of two expressions that,
Appear in the same form in both the equations. For example, you would be able to reduce to linear form two equations in x and y expressed in terms of
1
x
and
1
y
that appear in both equations,
Are inherently linear when inverted if originally inverted. For example, (x+y) is linear when appearing originally in both the equations as
1
x+y
, and
Appear unrelated to and independent of each other (not as a product or division). For example, in case any xy or
x
y
is present in any of the two equations, the equations can't be reduced to linear form by dummy variable substitution.
In the problem we have just solved, the two given equations were in terms of expressions
1
x
and
1
y
.
The two expressions are linear in x and y when inverted, appear unchanged in both equations and appear independently not as a product or division.
Let us solve a second problem to consolidate the new concept.
Example Problem 2.
Find the value of x and y by solving the following pair of equations algebraically.
3
3x+5y
−
11
x+2y
=−
9
10
and
5
3x+5y
+
1
x+2y
=
17
66
.
Solution to Example Problem 2—Solution of a pair non-linear equations by reducing to linear form with dummy variable substitution
The two equations are in terms of two expressions
1
3x+5y
and
1
x+2y
both of which satisfy the conditions for the equations to be reducible to linear form.
Substitute dummy variables p and q for these two expressions,
p=
1
3x+5y
, and,
q=
1
x+2y
.
The two given equations are transformed to,
3p−11q=−
9
10
, and
5p+q=
17
66
.
Decide variable q to be eliminated by multiplying the second equation with 11 and adding the result to the first equation. As a result of multiplication by 11 the second equation is changed to,
55p+11q=
17
6
.
Add it to the first equation to eliminate q. The result is,
58p=
17
6
−
9
10
=
58
30
,
Or, p=
1
30
=
1
3x+5y
.
So 3x+5y=30.
Substitute value of p in first equation. Result is,
1
10
−11q=−
9
10
,
Or, 11q=
1
10
+
9
10
=1,
Or, q=
1
11
=
1
x+2y
.
So, x+2y=11.
Solution to example problem 2: Second stage of solving two linear equations in x and y
You have now two linear equations in x and y,
3x+5y=30, and
x+2y=11.
Multiply the second equation by 3 to get,
3x+6y=33.
Subtract first equation from this result to get,
y=3.
Substitute value of y in the second equation. Result is,
x+6=11,
Or, x=5.
Answer: x=5, y=3.
