How do you solve the simultaneous equations \[y=3x-5\] and \[{{x}^{2}}+{{y}^{2}}=25\] ?
Answer
620.4k+ views
Hint: These types of problems are pretty straight forward and are very easy to solve. For problems like these, we must have a solid and firm understanding of the theory of simultaneous equations. In such problems, what we do is find the values of ‘x’ and ‘y’. First of all we find the value of an unknown parameter in terms of another unknown parameter and then put the value of this parameter in the other equation to find the value of the unknown parameter. We then put this value in any of the given equations to find the value of the other unknown parameter.
Complete step by step solution:
Now we start off with the solution to the problem by writing the two equations as,
\[y=3x-5\] ---- Equation \[1\]
\[{{x}^{2}}+{{y}^{2}}=25\] ----- Equation \[2\]
We now substitute the value of the first equation in the second equation and we get,
\[{{x}^{2}}+{{\left( 3x-5 \right)}^{2}}=25\]
Now, evaluating the value of the first unknown parameter, we get,
\[\begin{align}
& {{x}^{2}}+{{\left( 3x-5 \right)}^{2}}=25 \\
& \Rightarrow {{x}^{2}}+9{{x}^{2}}+25-30x=25 \\
& \Rightarrow 10{{x}^{2}}-30x=0 \\
\end{align}\]
Now taking \[x\] as common from the left hand side of the equation we get,
\[x\left( 10x-30 \right)=0\]
Now, from this we can write,
Either \[x=0\] or
\[\begin{align}
& 10x-30=0 \\
& \Rightarrow x=3 \\
\end{align}\]
Now, for \[x=0\], we get \[y=-5\] , by putting the value of \[x\] in the first equation and we get \[y=\pm 5\] by putting the value of \[x\] in the second equation.
Now, for \[x=3\] , we get \[y=4\] , by putting the value of \[x\] in the first equation and we get \[y=\pm 4\] by putting the value of \[x\] in the second equation.
Thus the solution to our problem are \[\left( x,y \right)=\left( 0,5 \right)\] and \[\left( x,y \right)=\left( 3,4 \right)\]
Note: For these kinds of problems, we need to be through with the chapter of simultaneous equations and substitution. These problems can also be done using graphs. We plot the graphs of both the functions. After plotting, we need to find the intersection points of both the graphs. These points of intersections are basically the solution to the problem, which give us the required values of ‘x’ and ‘y’.
Complete step by step solution:
Now we start off with the solution to the problem by writing the two equations as,
\[y=3x-5\] ---- Equation \[1\]
\[{{x}^{2}}+{{y}^{2}}=25\] ----- Equation \[2\]
We now substitute the value of the first equation in the second equation and we get,
\[{{x}^{2}}+{{\left( 3x-5 \right)}^{2}}=25\]
Now, evaluating the value of the first unknown parameter, we get,
\[\begin{align}
& {{x}^{2}}+{{\left( 3x-5 \right)}^{2}}=25 \\
& \Rightarrow {{x}^{2}}+9{{x}^{2}}+25-30x=25 \\
& \Rightarrow 10{{x}^{2}}-30x=0 \\
\end{align}\]
Now taking \[x\] as common from the left hand side of the equation we get,
\[x\left( 10x-30 \right)=0\]
Now, from this we can write,
Either \[x=0\] or
\[\begin{align}
& 10x-30=0 \\
& \Rightarrow x=3 \\
\end{align}\]
Now, for \[x=0\], we get \[y=-5\] , by putting the value of \[x\] in the first equation and we get \[y=\pm 5\] by putting the value of \[x\] in the second equation.
Now, for \[x=3\] , we get \[y=4\] , by putting the value of \[x\] in the first equation and we get \[y=\pm 4\] by putting the value of \[x\] in the second equation.
Thus the solution to our problem are \[\left( x,y \right)=\left( 0,5 \right)\] and \[\left( x,y \right)=\left( 3,4 \right)\]
Note: For these kinds of problems, we need to be through with the chapter of simultaneous equations and substitution. These problems can also be done using graphs. We plot the graphs of both the functions. After plotting, we need to find the intersection points of both the graphs. These points of intersections are basically the solution to the problem, which give us the required values of ‘x’ and ‘y’.
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