Question

How do you solve \[5=y-x\] and \[4{{x}^{2}}=-17x+y+4\]?

Answer 1

Hint: In this problem, we have to solve and find the value of x and y. We have two equations, where one equation can be substituted in another to solve and find the value of x and y. We can change the first equation to x or y equals, and substitute the value of x or y equals in the second equation. In the second equation, we will get a quadratic equation, from which we can find the x value and substituting the x value in the first equation to get the value of y.

Complete step by step answer:
We know that the given two equations are,
\[5=y-x\] ……. (1)
\[4{{x}^{2}}=-17x+y+4\] …….. (2)
Now we can write the equation (1) as
\[5+x=y\]
Now we can substitute the above value of y in equation (2), we get
\[\begin{align}
  & \Rightarrow 4{{x}^{2}}=-17x+x+5+4 \\
 & \Rightarrow 4{{x}^{2}}+16x-9=0.....(3) \\
\end{align}\]
Now we can use the quadratic formula for the above equation, we get
We know that the quadratic formula for the equation \[a{{x}^{2}}+bx+c=0\] is,
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
 Now we can compare the equation (3) with general equation, we get
a = 4, b = 16, c = -9
we can substitute the values in the quadratic formula, we get
\[\Rightarrow x=\dfrac{-16\pm \sqrt{{{\left( 16 \right)}^{2}}-4\left( 4 \right)\left( -9 \right)}}{2\left( 4 \right)}\]
We can now simplify the above step to get the value of x.
\[\begin{align}
  & \Rightarrow x=\dfrac{-16\pm \sqrt{256+144}}{8} \\
 & \Rightarrow x=\dfrac{-16\pm \sqrt{400}}{8} \\
 & \Rightarrow x=\dfrac{-16\pm \sqrt{{{\left( 20 \right)}^{2}}}}{8} \\
 & \Rightarrow x=\dfrac{-16\pm 20}{8} \\
\end{align}\]
Now we can split the terms, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-16+20}{8},\dfrac{-16-20}{8} \\
 & \Rightarrow x=\dfrac{4}{8},\dfrac{-36}{8} \\
 & \Rightarrow x=\dfrac{1}{2},\dfrac{-9}{2} \\
\end{align}\]
The value of \[x=\dfrac{1}{2},\dfrac{-9}{2}\].
Now we can substitute the x values in equation (1)
For x = \[\dfrac{1}{2}\] , the value of y is
\[\begin{align}
  & \Rightarrow y=5+\dfrac{1}{2} \\
 & \Rightarrow y=\dfrac{11}{2} \\
\end{align}\]
For x = \[\dfrac{-9}{2}\], the value of y is
\[\begin{align}
  & \Rightarrow y=5-\dfrac{9}{2} \\
 & \Rightarrow y=\dfrac{1}{2} \\
\end{align}\]
Therefore, the value of \[y=\dfrac{11}{2},\dfrac{1}{2}\]
 Therefore, The value of \[x=\dfrac{1}{2},\dfrac{-9}{2}\] and the value of \[y=\dfrac{11}{2},\dfrac{1}{2}\].

Note:
Students make mistakes while substituting the value of one equation in the other equation, which should be concentrated. We should know the quadratic formula, which we have used above, to solve and find the value of x. We can substitute the x value in any given equation to get the value of y.