Answer
405.3k+ views
Hint: We have a given set of equation:
$\begin{align}
& 3x+2y+25=0......(1) \\
& 2x+y+10=0......(2) \\
\end{align}$
To get the values of x and y we need to solve the given set of linear equations. We can solve a set of equations using a substitution method. In the substitution method, we get the value of one variable in terms of another variable from one equation and substitute it in the other equation to get the value of another variable. So, we will rearrange equation 1 and take y in terms of x and substitute it in rearranged equation 2.
Complete step-by-step solution
We can write the set of linear equations as:
$\begin{align}
& 3x+2y=-25......(3) \\
& 2x+y=-10......(4) \\
\end{align}$
So, by applying substitution method for the given set of linear equations:
From equation (4), we can write:
$\begin{align}
& 3x+2y=-25 \\
& \Rightarrow 2y=-25-3x \\
& \Rightarrow y=\dfrac{-25-3x}{2}......(5) \\
\end{align}$
Now, we get the value of y in terms of x.
Put the value of y in equation (3), we get:
$\begin{align}
& \Rightarrow 2x+y=-10 \\
& \Rightarrow 2x+\left( \dfrac{-25-3x}{2} \right)=-10 \\
& \Rightarrow 4x-25-3x=-20 \\
& \Rightarrow x=5......(6) \\
\end{align}$
Put the value of x from equation (6) in equation (5) to get the value of y:
So, we have:
\[\begin{align}
& y=\dfrac{-25-3\left( 5 \right)}{2} \\
& =\dfrac{-25-15}{2} \\
& =\dfrac{-40}{2} \\
& =-20
\end{align}\]
Hence, x = 5 and y = -20
Note: There is another method to solve a given set of linear equations, i.e. Elimination method. In the elimination method, we multiply or divide both the equations with a number to eliminate one variable in both the equations by adding or subtracting them and get the value of another value. Then we put the value in one of the equations and get the value of the eliminated variable.
For the given set of equations:
$\begin{align}
& 3x+2y=-25......(1) \\
& 2x+y=-10......(2) \\
\end{align}$
Multiply equation (1) by 2 and equation (2) by 3, we get:
$\begin{align}
& 6x+4y=-50......(3) \\
& 6x+3y=-30......(4) \\
\end{align}$
Subtract equation (4) from equation (3); we get:
$\begin{align}
& \Rightarrow 4y-3y=-50+30 \\
& \Rightarrow y=-20......(5) \\
\end{align}$
Now put the value of y from equation (5) in any of the above equation to get the value of x.
Let’s put value of y in equation (1), we get:
$\begin{align}
& \Rightarrow 3x+2y=-25 \\
& \Rightarrow 3x+2\left( -20 \right)=-255 \\
& \Rightarrow 3x-40=-25 \\
& \Rightarrow 3x=15 \\
& \Rightarrow x=5 \\
\end{align}$
$\begin{align}
& 3x+2y+25=0......(1) \\
& 2x+y+10=0......(2) \\
\end{align}$
To get the values of x and y we need to solve the given set of linear equations. We can solve a set of equations using a substitution method. In the substitution method, we get the value of one variable in terms of another variable from one equation and substitute it in the other equation to get the value of another variable. So, we will rearrange equation 1 and take y in terms of x and substitute it in rearranged equation 2.
Complete step-by-step solution
We can write the set of linear equations as:
$\begin{align}
& 3x+2y=-25......(3) \\
& 2x+y=-10......(4) \\
\end{align}$
So, by applying substitution method for the given set of linear equations:
From equation (4), we can write:
$\begin{align}
& 3x+2y=-25 \\
& \Rightarrow 2y=-25-3x \\
& \Rightarrow y=\dfrac{-25-3x}{2}......(5) \\
\end{align}$
Now, we get the value of y in terms of x.
Put the value of y in equation (3), we get:
$\begin{align}
& \Rightarrow 2x+y=-10 \\
& \Rightarrow 2x+\left( \dfrac{-25-3x}{2} \right)=-10 \\
& \Rightarrow 4x-25-3x=-20 \\
& \Rightarrow x=5......(6) \\
\end{align}$
Put the value of x from equation (6) in equation (5) to get the value of y:
So, we have:
\[\begin{align}
& y=\dfrac{-25-3\left( 5 \right)}{2} \\
& =\dfrac{-25-15}{2} \\
& =\dfrac{-40}{2} \\
& =-20
\end{align}\]
Hence, x = 5 and y = -20
Note: There is another method to solve a given set of linear equations, i.e. Elimination method. In the elimination method, we multiply or divide both the equations with a number to eliminate one variable in both the equations by adding or subtracting them and get the value of another value. Then we put the value in one of the equations and get the value of the eliminated variable.
For the given set of equations:
$\begin{align}
& 3x+2y=-25......(1) \\
& 2x+y=-10......(2) \\
\end{align}$
Multiply equation (1) by 2 and equation (2) by 3, we get:
$\begin{align}
& 6x+4y=-50......(3) \\
& 6x+3y=-30......(4) \\
\end{align}$
Subtract equation (4) from equation (3); we get:
$\begin{align}
& \Rightarrow 4y-3y=-50+30 \\
& \Rightarrow y=-20......(5) \\
\end{align}$
Now put the value of y from equation (5) in any of the above equation to get the value of x.
Let’s put value of y in equation (1), we get:
$\begin{align}
& \Rightarrow 3x+2y=-25 \\
& \Rightarrow 3x+2\left( -20 \right)=-255 \\
& \Rightarrow 3x-40=-25 \\
& \Rightarrow 3x=15 \\
& \Rightarrow x=5 \\
\end{align}$
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)