Question

How do you solve \[2x-4y=4\] and \[x+4y=14?\]

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 you can see that \[2x-4y=4\] and \[x+4y=14\] are linear equations.
\[2x-4y=4............(i)\]
\[x+4y=14............(ii)\]
Firstly, solve the second equation for \['x',\]
Step:-1
\[\Rightarrow x+4y=14\]
Subtract \['-4y'\] on both sides,
Therefore,
\[\Rightarrow x+4y-4y=14-4y\]
Above \['+4y-4y'\] will get canceled by each other.
Therefore,
\[\Rightarrow x=14-4y\]
Step:-2
Now, substitute \[\left( 14-4y \right)\] for the value of \[x\] in the first equation and solve for \[y:\]
\[\Rightarrow 2x-4y=4\] becomes,
\[\Rightarrow 2\left( 14-4y \right)-4y=4\]
\[\Rightarrow (2.14)(2.4y)-4y=4\]
\[\Rightarrow 28-8y-4y=4\]
Take common term from \[-8y-4y,\]
\[\Rightarrow 28+\left( -8-4 \right)y=4\]
\[\Rightarrow 28+\left( -12 \right)y=4\]
\[\Rightarrow 28-12y=4\]
Here subtract \['-28'\] on both sides,
\[-28+28-12y=4-28\]
Above \['-28+28'\] will get canceled by each other.
Therefore,
\[\Rightarrow -12y=4-28\]
\[\Rightarrow 12y=-24\]
Divide both the sides by \['-12',\]
\[\Rightarrow \dfrac{-12}{-12}=\dfrac{-24}{-12}\]
\[\Rightarrow y=2\]
Value of \[y\] is \[2.\]
Now, substitute the value of \[y\] as \[2\] in the solution of second equation at the end of step \[1\] and calculate the value of \[x,\]
\[\Rightarrow x=14-4y\] becomes
\[\Rightarrow x=14-\left( 4.2 \right)\]
\[\Rightarrow x=14-8\]
\[\Rightarrow x=6\]

Here the solution of \[2x-4y=4\] and \[x+4y=14\] is \[x=6\] and \[y=2\].

Note: In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables.
Linear equations are a combination of constants and variables.
The standard form 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.