
Solve the following pair of linear equations by the substitution method. \[3x-y=3\] and \[9x-3y=9\].
Answer
524.1k+ views
Hint: For a given equations of two lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\]and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\], if \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\] then they are parallel to each other. If two parallel lines have an infinite number of points common, then they are coincident lines. Coincident lines represent the same lines.
Complete step-by-step solution:
From the question it is clear that we have to solve a pair of linear equations \[3x-y=3\] and \[9x-3y=9\]. by the substitution method.
Consider
\[\Rightarrow 3x-y=3\]………………(1)
\[\Rightarrow 9x-3y=9\]…………….(2)
Now let us try to compare equation (1) with \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\]
\[{{a}_{1}}=3,{{b}_{1}}=-1,{{c}_{1}}=3\]
Now let us try to compare equation (2) with \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]
\[{{a}_{2}}=9,{{b}_{2}}=-3,{{c}_{2}}=9\]
Now we will check the condition \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]to verify whether these lines are parallel or not.
\[\Rightarrow \dfrac{3}{9}=\dfrac{-1}{-3}=\dfrac{3}{9}\]
After simplification we get
\[\Rightarrow \dfrac{1}{3}=\dfrac{1}{3}=\dfrac{1}{3}\]
So, we can conclude that these lines are parallel.
Now let us check whether these lines are coincident lines or not.
From the equation \[3x-y=3\] we can write \[y=3x-3\]
Now put \[y=3x-3\] in equation \[9x-3y=9\].
Now we can write
\[\Rightarrow 9x-3y=9\]
\[\Rightarrow 9x-3\left( 3x-3 \right)=9\]
\[\Rightarrow 9x-9x+9=9\]
After simplification we get,
\[\Rightarrow 9=9\]
\[9=9\] is true for any values of \[x\] and \[y\].
So, the lines \[3x-y=3\],\[9x-3x=9\] have infinitely many solutions and these lines are coincident lines.
Now let us try find one solution for both the lines
Now put \[y=0\] in \[3x-y=3\]
\[\Rightarrow 3x-y=3\]
\[\Rightarrow 3x-0=3\]
\[\Rightarrow 3x=3\]
After simplification we get
\[\Rightarrow x=1\]
At \[y=0\],\[x=1\]
\[\left( 1,0 \right)\] is one of the solutions of lines \[3x-y=3\],\[9x-3x=9\].
Since they are coincident lines, they represent the same line.
Note:Since\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\], many students may think that given lines are parallel so they do not have any solutions in common. But in some situations parallel lines can also be coincident lines.
In this case we have to check whether these lines are coincident lines. Otherwise students will do this question wrong.
Complete step-by-step solution:
From the question it is clear that we have to solve a pair of linear equations \[3x-y=3\] and \[9x-3y=9\]. by the substitution method.
Consider
\[\Rightarrow 3x-y=3\]………………(1)
\[\Rightarrow 9x-3y=9\]…………….(2)
Now let us try to compare equation (1) with \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\]
\[{{a}_{1}}=3,{{b}_{1}}=-1,{{c}_{1}}=3\]
Now let us try to compare equation (2) with \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]
\[{{a}_{2}}=9,{{b}_{2}}=-3,{{c}_{2}}=9\]
Now we will check the condition \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]to verify whether these lines are parallel or not.
\[\Rightarrow \dfrac{3}{9}=\dfrac{-1}{-3}=\dfrac{3}{9}\]
After simplification we get
\[\Rightarrow \dfrac{1}{3}=\dfrac{1}{3}=\dfrac{1}{3}\]
So, we can conclude that these lines are parallel.
Now let us check whether these lines are coincident lines or not.
From the equation \[3x-y=3\] we can write \[y=3x-3\]
Now put \[y=3x-3\] in equation \[9x-3y=9\].
Now we can write
\[\Rightarrow 9x-3y=9\]
\[\Rightarrow 9x-3\left( 3x-3 \right)=9\]
\[\Rightarrow 9x-9x+9=9\]
After simplification we get,
\[\Rightarrow 9=9\]
\[9=9\] is true for any values of \[x\] and \[y\].
So, the lines \[3x-y=3\],\[9x-3x=9\] have infinitely many solutions and these lines are coincident lines.
Now let us try find one solution for both the lines
Now put \[y=0\] in \[3x-y=3\]
\[\Rightarrow 3x-y=3\]
\[\Rightarrow 3x-0=3\]
\[\Rightarrow 3x=3\]
After simplification we get
\[\Rightarrow x=1\]
At \[y=0\],\[x=1\]
\[\left( 1,0 \right)\] is one of the solutions of lines \[3x-y=3\],\[9x-3x=9\].
Since they are coincident lines, they represent the same line.
Note:Since\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\], many students may think that given lines are parallel so they do not have any solutions in common. But in some situations parallel lines can also be coincident lines.
In this case we have to check whether these lines are coincident lines. Otherwise students will do this question wrong.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

