
How do you solve $y=2x-1$ and $y=x+2$?
Answer
558.9k+ views
Hint: There are two unknowns and two equations to solve. We solve the problem through substitution where we take one value of $x$ from one equation and place it in the other equation. Then we verify the result by solving 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 answer:
The given equations $y=2x-1$ and $y=x+2$ 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 can also find the value of one variable y with respect to x based on the equation
$y=x+2$.
We will replace the value of y for the substitution process in the second equation of
$y=2x-1$ and get
\[
\Rightarrow y=2x-1 \\
\Rightarrow x+2=2x-1 \\
\Rightarrow 2x-x=2+1 \\
\Rightarrow x=3 \\
\]
The solution for the variable $x$ is 3.
We put the values in the equation $y=x+2$ to find the value of $y$.
So, $y=x+2=3+2=5$.
The solutions for the $y=2x-1$ and $y=x+2$ are $x=3;y=2$.
Note: The coefficients of the given equation define the number of solutions for the given equations.
Let’s take two equations and two unknowns. $ax+by=m$ and $cx+dy=n$
In the equations the terms a, b, c, d, m, n all are constants.
We take the ratios of the coefficients of the same variables. The ratios are $\dfrac{a}{c},\dfrac{b}{d},\dfrac{m}{n}$.
The relation among these ratios determines the equations.
If $\dfrac{a}{c}=\dfrac{b}{d}\ne \dfrac{m}{n}$, then we have no solution for those equations.
If $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{m}{n}$, then we have an infinite number of solutions for those equations.
If $\dfrac{a}{c}\ne \dfrac{b}{d}$, then we have exactly one solution for those equations.
Complete step by step answer:
The given equations $y=2x-1$ and $y=x+2$ 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 can also find the value of one variable y with respect to x based on the equation
$y=x+2$.
We will replace the value of y for the substitution process in the second equation of
$y=2x-1$ and get
\[
\Rightarrow y=2x-1 \\
\Rightarrow x+2=2x-1 \\
\Rightarrow 2x-x=2+1 \\
\Rightarrow x=3 \\
\]
The solution for the variable $x$ is 3.
We put the values in the equation $y=x+2$ to find the value of $y$.
So, $y=x+2=3+2=5$.
The solutions for the $y=2x-1$ and $y=x+2$ are $x=3;y=2$.
Note: The coefficients of the given equation define the number of solutions for the given equations.
Let’s take two equations and two unknowns. $ax+by=m$ and $cx+dy=n$
In the equations the terms a, b, c, d, m, n all are constants.
We take the ratios of the coefficients of the same variables. The ratios are $\dfrac{a}{c},\dfrac{b}{d},\dfrac{m}{n}$.
The relation among these ratios determines the equations.
If $\dfrac{a}{c}=\dfrac{b}{d}\ne \dfrac{m}{n}$, then we have no solution for those equations.
If $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{m}{n}$, then we have an infinite number of solutions for those equations.
If $\dfrac{a}{c}\ne \dfrac{b}{d}$, then we have exactly one solution for those equations.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

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

Which women's tennis player has 24 Grand Slam singles titles?

Who is the Brand Ambassador of Incredible India?

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

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

