Question

How do solve the following linear system\[?\]
\[-6x+6y=6,\,\,\,\,-6x+3y=-12?\]

Answer 1

Hint: A way to solve a linear system algebraically is to use the substitution method. The substitution method functions by substituting the one \[y\] value with the other. We are going to explain this by using an example. We can substitute \['y'\] in the second equation with the first equation since \[y=y.\]

Complete step by step solution:
As we know that, given linear system is
\[-6x+6y=6,\,\,\,\,and\,\,\,-6x+3y=-12\]
\[\,-6x+6y=6\,............(i)\]
\[\,-6x+3y=-12............(ii)\]
Subtract the equation \[{{1}^{st}}\] with equation \[{{2}^{nd}},\]
\[\,-6x+6y+6x-3y =6+12\]
\[\,0x+3y=18\]
Therefore,
\[3y=18\]
\[y=\dfrac{18}{3}\]
\[y=6\]
Put the value of \[y\] in equation \[\left( i \right),\]
\[-6\times x+6\times 6=6\]
\[-6x+36=6\]
\[-6x=6-36\]
\[-6x=-30\]
Here above \['-'\] sign on both sides will get cancel by each other,
Therefore,
\[6x=30\]
\[x=\dfrac{30}{6}\]
\[x=5\]

Hence by solving given linear system \[-6x+6y=6\] and \[-6x+3y=12\] the value of \[x\] is \[5\] \[y\] is \[6.\]

Note: Linear equations are a combination of contents and variables.
The standard term of a linear equation in one variable is represented as \[ax+b=0\] where \[a\ne 0\] means a cannot be equal to zero and \[x\]is the variable. \[ax+by+c=0\] where \[a\ne 0,\] \[b\ne 0,x\] and \[y\] are the variables.