
Solve the system by Substitution method. Show your work \[2y-x=5,\] \[{{x}^{2}}+{{y}^{2}}-25=0\]?
Answer
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Hint: In order to find the solution to the given question, that is to solve the system: \[2y-x=5,\] \[{{x}^{2}}+{{y}^{2}}-25=0\]by Substitution method. We need to find the value of one variable say \[x\] in terms of \[y\] from one of the given equations and substitute in the other equation. Like this, find the value of both variables\[x\] and \[y\].
Complete step-by-step answer:
According to the question, given system of equation in the question is as follows:
\[\begin{align}
& 2y-x=5...\left( 1 \right) \\
& {{x}^{2}}+{{y}^{2}}-25=0...\left( 2 \right) \\
\end{align}\]
Now, apply the substitution method and rearrange the equation \[\left( 1 \right)\] to give variable \[x\] in terms of \[y\]
\[x=2y-5...\left( 3 \right)\]
After this substitute the equation \[\left( 3 \right)\]into equation \[\left( 2 \right)\], we get:
\[\Rightarrow {{\left( 2y-5 \right)}^{2}}+{{y}^{2}}-25=0\]
Now expand the brackets and simplify the above expression with the help of the identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], we get:
\[\Rightarrow 4{{y}^{2}}-20y+25+{{y}^{2}}-25=0\]
Simplifying it further we get:
\[\Rightarrow 5{{y}^{2}}-20y=0\]
Now after this take out a common factor \[5y\] from the above expression in the left-hand side of the equation, we get:
\[\Rightarrow 5y\left( y-4 \right)=0\]
To solve it further, equate each factor to zero from the above equation and solve for \[y\], we get:
\[\begin{align}
& \Rightarrow 5y=0 \\
& \Rightarrow y=0 \\
& \Rightarrow y-4=0 \\
& \Rightarrow y=4 \\
\end{align}\]
Now substitute these values into equation \[\left( 3 \right)\] for corresponding values of \[x\], we get:
For \[y=0\]:
\[\Rightarrow x=0-5=-5\]
Therefore, the point is \[\left( -5,0 \right)\]
For \[y=4\]:
\[\Rightarrow x=8-5=3\]
Therefore, the point is \[\left( 3,4 \right)\]
Hence the points of intersection are \[\left( -5,0 \right)\] and \[\left( 3,4 \right)\].
Note: Students can apply the substitution method another way also like by finding the value of one variable say \[y\] in terms of \[x\] from one of the given equations and substitute in the other equation. Like this then find the value of both variables\[x\] and \[y\].
Complete step-by-step answer:
According to the question, given system of equation in the question is as follows:
\[\begin{align}
& 2y-x=5...\left( 1 \right) \\
& {{x}^{2}}+{{y}^{2}}-25=0...\left( 2 \right) \\
\end{align}\]
Now, apply the substitution method and rearrange the equation \[\left( 1 \right)\] to give variable \[x\] in terms of \[y\]
\[x=2y-5...\left( 3 \right)\]
After this substitute the equation \[\left( 3 \right)\]into equation \[\left( 2 \right)\], we get:
\[\Rightarrow {{\left( 2y-5 \right)}^{2}}+{{y}^{2}}-25=0\]
Now expand the brackets and simplify the above expression with the help of the identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], we get:
\[\Rightarrow 4{{y}^{2}}-20y+25+{{y}^{2}}-25=0\]
Simplifying it further we get:
\[\Rightarrow 5{{y}^{2}}-20y=0\]
Now after this take out a common factor \[5y\] from the above expression in the left-hand side of the equation, we get:
\[\Rightarrow 5y\left( y-4 \right)=0\]
To solve it further, equate each factor to zero from the above equation and solve for \[y\], we get:
\[\begin{align}
& \Rightarrow 5y=0 \\
& \Rightarrow y=0 \\
& \Rightarrow y-4=0 \\
& \Rightarrow y=4 \\
\end{align}\]
Now substitute these values into equation \[\left( 3 \right)\] for corresponding values of \[x\], we get:
For \[y=0\]:
\[\Rightarrow x=0-5=-5\]
Therefore, the point is \[\left( -5,0 \right)\]
For \[y=4\]:
\[\Rightarrow x=8-5=3\]
Therefore, the point is \[\left( 3,4 \right)\]
Hence the points of intersection are \[\left( -5,0 \right)\] and \[\left( 3,4 \right)\].
Note: Students can apply the substitution method another way also like by finding the value of one variable say \[y\] in terms of \[x\] from one of the given equations and substitute in the other equation. Like this then find the value of both variables\[x\] and \[y\].
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