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How do you solve x+y=6 and x-y=9?

Answer
VerifiedVerified
564.3k+ views
Hint: Any two linear equations can be solved to get a common point lying on their graphs. We solve these equations by addition and subtraction methods. We can solve these equations by adding them after making required modifications such that any one of the variables gets canceled. Then, we get the value of one variable. Using this we get the other variable value.

Complete step-by-step solution:
As per the given question we need to solve \[x+y=6\] and \[x-y=9\] to get a common point called a solution of these equations.
If we see carefully, we observe that the sign of y in both equations is the opposite. And the coefficient of y term in both the equations is equal. So, we need not multiply the given equations.
Now, let us add both the equations \[x+y=6\] and \[x-y=9\]. That is, we write it as
\[\Rightarrow (x+y)+(x-y)=(6+9)\]
On rearranging the terms by keeping similar terms together, we get it as
\[\Rightarrow x+x+y-y=6+9\]
The addition of x and x gives \[2x\] and subtraction of y from y gives nothing. That is y term gets eliminated in the above equation. And, in addition to 6 and 9, we get 15.
Then, we get
\[\Rightarrow 2x=15\]
\[\therefore x=\dfrac{15}{2}\] or \[x=7.5\] is the value of x at the intersection of the given two lines.
Now, let us substitute the value of x, \[x=7.5\]into the equation \[x+y=6\]. That is, we can write
\[\Rightarrow (7.5)+y=6\]
The equation can be rewritten as shown below,
\[\Rightarrow y=6-7.5\]
Subtracting 7.5 from 6, we get
\[\Rightarrow y=-1.5\]
\[\therefore \] x = 7.5 and y=-1.5 is the required solution.

Note: We can solve for x and y of two equations by substituting \[x=6-y\] in the second equation \[x-y=9\] to get the y value and substituting the obtained y value in \[x=6-y\] to get the respective x value. In general, we write one variable in terms of the other by rearranging the terms in any one of the two equations. Then, we substitute the rearranged variable into the second equation and find the value of each variable.