
How do you solve the system of linear equations $6u+v=18$ and $5u+2v=22$?
Answer
541.5k+ views
Hint: There are two unknowns $u$ and $v$ and also two equations to solve. 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. First, we are applying the process of reduction and then the substitution.
Complete step-by-step solution:
The given equations $6u+v=18$ and $5u+2v=22$ 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 $6u+v=18.....(i)$ and $5u+2v=22......(ii)$.
We multiply 2 to the both sides of the first equation and get
$\begin{align}
& 2\times \left( 6u+v \right)=18\times 2 \\
& \Rightarrow 12u+2v=36 \\
\end{align}$
We take the equation as $12u+2v=36.....(iii)$.
Now we subtract the equation (ii) from equation (iii) and get
$\left( 12u+2v \right)-\left( 5u+2v \right)=36-22$.
We take the variables together and the constants on the other side.
Simplifying the equation, we get
$\begin{align}
& \left( 12u+2v \right)-\left( 5u+2v \right)=36-22 \\
& \Rightarrow 7u=14 \\
& \Rightarrow u=\dfrac{14}{7}=2 \\
\end{align}$
The value of $u$ is 2. Now putting the value in the equation $6u+v=18.....(i)$, we get
$\begin{align}
& 6u+v=18 \\
& \Rightarrow v=18-6\times 2=6 \\
\end{align}$.
Therefore, the values are $u=2,v=6$.
Note: We can also find the value of one variable $v$ with respect to $u$ based on the equation
$6u+v=18$ where $v=18-6u$. We replace the value of $v$ in the second equation of
$5u+2v=22$ and get
\[\begin{align}
& 5u+2v=22 \\
& \Rightarrow 5u+2\left( 18-6u \right)=22 \\
& \Rightarrow 5u+36-12u=22 \\
\end{align}\]
We get the equation of $u$ and solve
$\begin{align}
& 5u+36-12u=22 \\
& \Rightarrow -7u=22-36=-14 \\
& \Rightarrow u=\dfrac{-14}{-7}=2 \\
\end{align}$
Putting the value of $u$ we get $v=18-6u=18-6\times 2=6$.
Therefore, the values are $u=2,v=6$.
Complete step-by-step solution:
The given equations $6u+v=18$ and $5u+2v=22$ 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 $6u+v=18.....(i)$ and $5u+2v=22......(ii)$.
We multiply 2 to the both sides of the first equation and get
$\begin{align}
& 2\times \left( 6u+v \right)=18\times 2 \\
& \Rightarrow 12u+2v=36 \\
\end{align}$
We take the equation as $12u+2v=36.....(iii)$.
Now we subtract the equation (ii) from equation (iii) and get
$\left( 12u+2v \right)-\left( 5u+2v \right)=36-22$.
We take the variables together and the constants on the other side.
Simplifying the equation, we get
$\begin{align}
& \left( 12u+2v \right)-\left( 5u+2v \right)=36-22 \\
& \Rightarrow 7u=14 \\
& \Rightarrow u=\dfrac{14}{7}=2 \\
\end{align}$
The value of $u$ is 2. Now putting the value in the equation $6u+v=18.....(i)$, we get
$\begin{align}
& 6u+v=18 \\
& \Rightarrow v=18-6\times 2=6 \\
\end{align}$.
Therefore, the values are $u=2,v=6$.
Note: We can also find the value of one variable $v$ with respect to $u$ based on the equation
$6u+v=18$ where $v=18-6u$. We replace the value of $v$ in the second equation of
$5u+2v=22$ and get
\[\begin{align}
& 5u+2v=22 \\
& \Rightarrow 5u+2\left( 18-6u \right)=22 \\
& \Rightarrow 5u+36-12u=22 \\
\end{align}\]
We get the equation of $u$ and solve
$\begin{align}
& 5u+36-12u=22 \\
& \Rightarrow -7u=22-36=-14 \\
& \Rightarrow u=\dfrac{-14}{-7}=2 \\
\end{align}$
Putting the value of $u$ we get $v=18-6u=18-6\times 2=6$.
Therefore, the values are $u=2,v=6$.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

What happens to glucose which enters nephron along class 10 biology CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which of the following does not have a fundamental class 10 physics CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

