Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How do you solve this system of equations \[11x+15y+23=0\] and \[7x-2y-20=0\]?

seo-qna
Last updated date: 20th Jun 2024
Total views: 374.4k
Views today: 9.74k
Answer
VerifiedVerified
374.4k+ views
Hint: To solve the system of equations in two variables, we need to follow the steps given below in the same order:
Step 1: choose one of the equations to find the relationship between the two variables. This can be done by taking one of the variables to the other side of the equation.
Step 2: substitute this relationship in the other equation to get an equation in one variable.
Step 3: solve this equation to find the solution value of the variable.
Step 4: substitute this value in any of the equations to find the value of the other variable.

Complete step by step solution:
We are given the two equations \[11x+15y+23=0\] and \[7x-2y-20=0\]. We know the steps required to solve a system of equations in two variables. Let’s take the first equation, we get
\[\Rightarrow 11x+15y+23=0\]
Subtracting \[15y\] from both sides of equation, we get
\[\Rightarrow 11x+23=-15y\]
Dividing both sides by \[-15\] and flipping the sides, we get
\[\Rightarrow y=\dfrac{11x+23}{-15}\]
Substituting this in the equation \[7x-2y-20=0\], we get
\[\Rightarrow 7x-2\left( \dfrac{11x+23}{-15} \right)-20=0\]
Simplifying the above equation, we get
\[\begin{align}
  & \Rightarrow 105x+22x+46-300=0 \\
 & \Rightarrow 127x-254=0 \\
\end{align}\]
Solving the above equation, we get
\[\Rightarrow x=2\]
Substituting this value in the relationship between variables to find the value of y, we get
\[\Rightarrow y=\dfrac{11(2)+23}{-15}=\dfrac{45}{-15}=-3\]
Hence, the solution values for the system of equations are \[x=2\And y=-3\].

Note:
To solve any system of equations having two variables, we need to follow the given steps. Unlike this question, even if the degrees of the equations are different. It should be noted that we must check that both functions are defined on that value or not.