Question

How do you solve this system of equations: \[6x + y = 7\], \[y + z = 2\] and \[4x - z = 4\]

Answer 1

Hint: In this question we have to find the unknown values that are \[x\], \[y\] and \[z\] solving these equations by using the elimination method. In elimination methods either we add or subtract the equations to find the unknown. hence we obtain the required result.

Complete step by step solution:
Here we have 3 equations. So now we consider the 3 equations.
\[\Rightarrow 6x + y = 7\]-------(1)
\[\Rightarrow y + z = 2\]--------(2)
\[\Rightarrow 4x - z = 4\]-------(3)
Now we will subtract the equation (2) from the equation (1) so we have
\[\Rightarrow 6x + y - (y + z) = 7 - 2\]
On simplifying we have
\[\Rightarrow 6x + y - y - z = 7 - 2\]
The y and -y will get cancelled. We have
\[\Rightarrow 6x - z = 5\]-------(4)
Now we will consider equation (3) and equation (4) because the two equations contain the same variable.
\[

   + 4x - z = 4 \\
  \underline{ \mathop + \limits_{( - )} 6x\mathop - \limits_{( + )} z = \mathop + \limits_{( - )} 5} \\
    \\
   - 2x = - 1 \\
 \]
The negative sign will get canceled. so we have
\[ \Rightarrow 2x = 1\]
On dividing by 2 we have
\[ \Rightarrow x = \dfrac{1}{2}\]
Hence we have determined the value of x and now we have to determine the value of y and z.
Now substitute the value of x in the equation (1) we have
\[ \Rightarrow 6\left( {\dfrac{1}{2}} \right) + y = 7\]
On simplifying we have
\[ \Rightarrow 3 + y = 7\]
Take 3 to RHS we have
\[ \Rightarrow y = 7 - 3\]
On further simplifying we have
\[ \Rightarrow y = 4\]
Hence we have determined the value of y and now we have to determine the value of z.
Now substitute the value of y in the equation (2) we have
\[ \Rightarrow 4 + z = 2\]
Take 4 to RHS we have
\[ \Rightarrow z = 2 - 4\]
On further simplifying we have
\[ \Rightarrow z = - 2\]
Hence we have determined the value of z.

Note: In this type of question while eliminating the term we must be aware of the sign where we change the sign by the alternate sign. In this we have a chance to verify our answers. In the elimination method we have made the one term have the same coefficient such that it will be easy to solve the equation.