
The cost of four chairs and five tables is Rs 3200. Write a linear equation in two variables for this statement and find out its two solutions.
Answer
593.1k+ views
Hint: First we will first assume that \[x\] be the cost of the chair and \[y\] be the cost of the table and then write the linear equation from the given condition. Then we will choose values of \[x\] and \[y\] arbitrarily to find the two solutions of the obtained equation.
Complete step by step answer:
We are given that the cost of four chairs and five tables is Rs 3200.
Let us assume that \[x\] be the cost of the chair and \[y\] be the cost of the table.
Writing the linear equation from the given condition, we get
\[ \Rightarrow 4x + 5y = 3200{\text{ ......eq.(1)}}\]
Replacing 5 for \[x\] randomly in the equation (1), we get
\[
\Rightarrow 4\left( 5 \right) + 5y = 3200 \\
\Rightarrow 20 + 5y = 3200 \\
\]
Subtracting the above equation by 20 on both sides, we get
\[
\Rightarrow 20 + 5y - 20 = 3200 - 20 \\
\Rightarrow 5y = 3180 \\
\]
Dividing the above equation by 5 on both sides, we get
\[
\Rightarrow \dfrac{{5y}}{5} = \dfrac{{3180}}{5} \\
\Rightarrow y = 636 \\
\]
Therefore, the first solution is \[\left( {5,636} \right)\].
Replacing 4 for \[y\] arbitrarily in the equation (1), we get
\[
\Rightarrow 4x + 5\left( 4 \right) = 3200 \\
\Rightarrow 4x + 20 = 3200 \\
\]
Subtracting the above equation by 20 on both sides, we get
\[
\Rightarrow 4x + 20 - 20 = 3200 - 20 \\
\Rightarrow 4x = 3180 \\
\]
Dividing the above equation by 4 on both sides, we get
\[
\Rightarrow \dfrac{{4x}}{4} = \dfrac{{3180}}{4} \\
\Rightarrow x = 795 \\
\]
Therefore, the second solution is \[\left( {795,4} \right)\].
Note: Students should know that there will be two variables since the given problem has only two items, that are, chairs and tables. One should know that a linear equation in two variable is an equation where there are two distinct variables \[x\] and \[y\] written in the form of \[ax + by = c\], where \[a\], \[b\] and \[c\] are real numbers.
Complete step by step answer:
We are given that the cost of four chairs and five tables is Rs 3200.
Let us assume that \[x\] be the cost of the chair and \[y\] be the cost of the table.
Writing the linear equation from the given condition, we get
\[ \Rightarrow 4x + 5y = 3200{\text{ ......eq.(1)}}\]
Replacing 5 for \[x\] randomly in the equation (1), we get
\[
\Rightarrow 4\left( 5 \right) + 5y = 3200 \\
\Rightarrow 20 + 5y = 3200 \\
\]
Subtracting the above equation by 20 on both sides, we get
\[
\Rightarrow 20 + 5y - 20 = 3200 - 20 \\
\Rightarrow 5y = 3180 \\
\]
Dividing the above equation by 5 on both sides, we get
\[
\Rightarrow \dfrac{{5y}}{5} = \dfrac{{3180}}{5} \\
\Rightarrow y = 636 \\
\]
Therefore, the first solution is \[\left( {5,636} \right)\].
Replacing 4 for \[y\] arbitrarily in the equation (1), we get
\[
\Rightarrow 4x + 5\left( 4 \right) = 3200 \\
\Rightarrow 4x + 20 = 3200 \\
\]
Subtracting the above equation by 20 on both sides, we get
\[
\Rightarrow 4x + 20 - 20 = 3200 - 20 \\
\Rightarrow 4x = 3180 \\
\]
Dividing the above equation by 4 on both sides, we get
\[
\Rightarrow \dfrac{{4x}}{4} = \dfrac{{3180}}{4} \\
\Rightarrow x = 795 \\
\]
Therefore, the second solution is \[\left( {795,4} \right)\].
Note: Students should know that there will be two variables since the given problem has only two items, that are, chairs and tables. One should know that a linear equation in two variable is an equation where there are two distinct variables \[x\] and \[y\] written in the form of \[ax + by = c\], where \[a\], \[b\] and \[c\] are real numbers.
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