Question

How do you solve the system of equations and check algebraically?
$y={{x}^{2}}-6x+1,y+2x=6$

Answer 1

Hint: Assume the given equations as equation (1) and (2) respectively. Now, find the value of y in terms of x from equation (2) and substitute this value in equation (1) to form a quadratic expression in x. Use the middle term split method to solve for the two values of x. Once the value of x is found, substitute it in equation (2) to get the respective two values of y. Now, to check if we got the correct set of points or not, substitute the coordinates of the two points obtained one – by – one in equation (1). If the point satisfies the equation then our answer is correct otherwise not.

Complete step by step answer:
Here, we have been provided with two equations: $y={{x}^{2}}-6x+1$ and $y+2x=6$, we are asked to solve this system and check algebraically. That means we have to find the values of the variables x and y and check them if they satisfy the given equations or not.
Let us assume the two given equations as equation (1) and equation (2) respectively, so we have,
$\Rightarrow y={{x}^{2}}-6x+1$ - (1)
\[\Rightarrow y+2x=6\] - (2)
Now, there two variables x and y so to solve the equation we need to change it into an expression containing one variable. In equation (2) we can write y in terms of x and then we can substitute it in equation (1). So we get,
\[\begin{align}
  & \Rightarrow 6-2x={{x}^{2}}-6x+1 \\
 & \Rightarrow {{x}^{2}}-4x-5=0 \\
\end{align}\]
Using the middle term split method to solve the above quadratic expression we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}-5x+x-5=0 \\
 & \Rightarrow \left( x-5 \right)\left( x+1 \right)=0 \\
\end{align}\]
Substituting each term equal to 0 we get,
\[\therefore x=5\] or \[x=-1\]
Now, substituting the obtained values of x in equation (2) we get,
(a) For x = 5 we have,
\[\Rightarrow y=-4\]
(b) For x = -1 we have,
\[\Rightarrow y=8\]
Therefore, the solution points are: (5, -4) and (-1, 8).
Now, we need to check if these points satisfy equation (1) or not. Let us consider them one by one.
(i) Considering the point (5, -4) we get,
$\begin{align}
  & \Rightarrow -4={{5}^{2}}-6\left( 5 \right)+1 \\
 & \Rightarrow -4=25-30+1 \\
 & \Rightarrow -4=-4 \\
\end{align}$
Therefore, the point (5, -4) satisfies the equation.
(ii) Considering the point (-1, 8) we get,
$\begin{align}
  & \Rightarrow 8={{\left( -1 \right)}^{2}}-6\left( -1 \right)+1 \\
 & \Rightarrow 8=1+6+1 \\
 & \Rightarrow 8=8 \\
\end{align}$
Therefore, the point (-1, 8) satisfies the equation.
Hence, we can conclude that both the points are the solutions of the given system of equations.

Note: Here you can also write x in terms of y from the second equation and then substitute it in the first equation to form a quadratic in y. Then you may solve for the value of y first. But do not select equation (1) to write the value of x in terms of y. If you will do so then you will get the radical sign which we will be required to remove while solving, that may take extra time.