
How do you solve the following linear system $4x-2y=2$ and $x-4y=4$ ?
Answer
554.7k+ views
Hint: There are 2 unknown variables and 2 equations in the given question. This is a linear equation in 2 variables. We can solve the equation by eliminating any one of the 2 variables.
Complete step-by-step answer:
The given 2 equations in the question
$4x-2y=2$ …..eq1
$x-4y=4$ ……eq2
Let’s solve the question by eliminating b.
In eq1 we can multiply 2 in both LHS and RHS
The equation will be $8x-4y=4$ ….eq3
Now subtracting eq2 from eq3
7x = 0
So x =0
Now we can get the value of y by substituting the value of x in any one of the equations.
Let’s put the value of x in eq2
$0-4y=4$
So y = -1
Now we can verify the answers by putting the value of x and y in both the equations.
Note: Another method is to solve by determinant method
Suppose there are 2 linear equation ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$ the solution to this problem let’s take
D= $\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} \\
{{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right|$
A= $\left| \begin{matrix}
{{b}_{1}} & {{c}_{1}} \\
{{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right|$
B= $\left| \begin{matrix}
{{a}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{c}_{2}} \\
\end{matrix} \right|$
$x=-\dfrac{A}{D}$ and $y=\dfrac{B}{D}$ Where $D\ne 0$
If $D=0$ and any one of A and B is not equal to 0 then there will be no solution for the system of equation
If $D=A=B=0$ there will be infinitely many solutions to the system of equations.
If $D\ne 0$ there will be one solution for the system of equations.
In this case ${{a}_{1}}=4,{{b}_{1}}=-2,{{c}_{1}}=2$ and ${{a}_{2}}=1,{{b}_{2}}=-4,{{c}_{2}}=4$
Solving the determinant $D=-14,A=0$ and $B=14$
So $x=-\dfrac{A}{D}$
So x = 0
And $y=\dfrac{B}{D}$= -1
While solving the system of linear equations we can imagine the equation having 2 unknown as equations of straight in 2-D Cartesian plane so 2 equations means 2 lines will intersect at 1 point unless both lines are parallel or both lines are overlapping. Same goes with 3 unknowns, we can imagine the equations as equations of plane in 3D geometry and then solve for the unknown variables.
Complete step-by-step answer:
The given 2 equations in the question
$4x-2y=2$ …..eq1
$x-4y=4$ ……eq2
Let’s solve the question by eliminating b.
In eq1 we can multiply 2 in both LHS and RHS
The equation will be $8x-4y=4$ ….eq3
Now subtracting eq2 from eq3
7x = 0
So x =0
Now we can get the value of y by substituting the value of x in any one of the equations.
Let’s put the value of x in eq2
$0-4y=4$
So y = -1
Now we can verify the answers by putting the value of x and y in both the equations.
Note: Another method is to solve by determinant method
Suppose there are 2 linear equation ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$ the solution to this problem let’s take
D= $\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} \\
{{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right|$
A= $\left| \begin{matrix}
{{b}_{1}} & {{c}_{1}} \\
{{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right|$
B= $\left| \begin{matrix}
{{a}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{c}_{2}} \\
\end{matrix} \right|$
$x=-\dfrac{A}{D}$ and $y=\dfrac{B}{D}$ Where $D\ne 0$
If $D=0$ and any one of A and B is not equal to 0 then there will be no solution for the system of equation
If $D=A=B=0$ there will be infinitely many solutions to the system of equations.
If $D\ne 0$ there will be one solution for the system of equations.
In this case ${{a}_{1}}=4,{{b}_{1}}=-2,{{c}_{1}}=2$ and ${{a}_{2}}=1,{{b}_{2}}=-4,{{c}_{2}}=4$
Solving the determinant $D=-14,A=0$ and $B=14$
So $x=-\dfrac{A}{D}$
So x = 0
And $y=\dfrac{B}{D}$= -1
While solving the system of linear equations we can imagine the equation having 2 unknown as equations of straight in 2-D Cartesian plane so 2 equations means 2 lines will intersect at 1 point unless both lines are parallel or both lines are overlapping. Same goes with 3 unknowns, we can imagine the equations as equations of plane in 3D geometry and then solve for the unknown variables.
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