Courses
Courses for Kids
Free study material
Offline Centres
More

How do you solve the system $2x+7y=3$ and $x=1-4y$ by substitution?

Last updated date: 01st Mar 2024
Total views: 341.7k
Views today: 3.41k
Verified
341.7k+ views
Hint: There are two unknowns $x$ and $y$ and also two equations to solve. We are applying the process of substitution and then the reduction. We take the value of the one variable and place that on another equation to solve the variables. We solve the equations equating the coefficients of one variable and omitting the variable. The other variable remains with the constants. Using the binary operation, we find the value of the other variable.

Complete step-by-step solution:
The given equations $2x+7y=3$ and $x=1-4y$ are linear equations of two variables.
We know that the number of equations has to be equal to the number of unknowns to solve them.
We take the equations as $2x+7y=3.....(i)$ and $x=1-4y......(ii)$.
We can also find the value of one variable $x$ with respect to $y$ based on the equation
$x=1-4y$. We replace the value of $x$ in the first equation of $2x+7y=3$ and get
\begin{align} & 2x+7y=3 \\ & \Rightarrow 2\left( 1-4y \right)+7y=3 \\ & \Rightarrow 2-8y+7y=3 \\ \end{align}
We get the equation of $y$ and solve
\begin{align} & \Rightarrow 2-8y+7y=3 \\ & \Rightarrow 2-y=3 \\ & \Rightarrow y=2-3=-1 \\ \end{align}
Putting the value of $x$ we get $x=1-4y=1-4\left( -1 \right)=1+4=5$.
Therefore, the values are $x=5,y=-1$.

Note: Now we solve it through the reduction method.
We multiply 2 to the both sides of the second equation $x=1-4y$ and get
\begin{align} & 2\times x=2\times \left( 1-4y \right) \\ & \Rightarrow 2x=2-8y \\ \end{align}
We subtract this equation from $2x+7y=3$ to get
\begin{align} & \left( 2x+7y \right)-2x=3-\left( 2-8y \right) \\ & \Rightarrow 7y=1+8y \\ & \Rightarrow y=-1 \\ \end{align}
The value of $y$ is $-1$. Now putting the value in the equation $2x+7y=3.....(i)$, we get
\begin{align} & 2x+7y=3 \\ & \Rightarrow 2x=3+7 \\ & \Rightarrow x=\dfrac{10}{2}=5 \\ \end{align}.
Therefore, the values are $x=5,y=-1$.