Champa went to a 'Sale' to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants she purchased". Help her friends to find how many pants and skirts Champa bbough
pls answer this question quickly pls
Answers & Comments
Answer:
[tex]\begin{gathered}\begin{gathered}\bf\: \begin{cases} &\bf{Number \: of \: pants \: purchased \: = \: 1} \\ \\ &\bf{Number \: of \: skirts \: purchased \: = \: 0} \end{cases}\end{gathered}\end{gathered}[/tex]
Explanation:
[tex]\begin{gathered}\begin{gathered}\bf\: Let\:assume\:that-\begin{cases} &\sf{number \: of \: pants \: purchased \: be \: x} \\ \\ &\sf{number \: of \: skirts \: purchased \: be \: y} \end{cases}\end{gathered}\end{gathered}[/tex]
Given that, When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased.''
[tex]\implies\bf \: y = 2x - 2 \: \cdots \: \cdots(1) \\ [/tex]
Further given that, the number of skirts is four less than four times the number of pants she purchased.
[tex]\implies\bf \: y = 4x - 4 \: \\ [/tex]
On substituting the value of y from equation (1), we get
[tex]\sf \: 2x - 2 = 4x - 4 \\ [/tex]
[tex]\sf \: 4x - 2x = 4 - 2 \\ [/tex]
[tex]\sf \: 2x = 2 \\ [/tex]
[tex]\implies\sf \: \boxed{\bf \: x = 1 \: } \\ [/tex]
On substituting x = 1, in equation (1), we get
[tex]\sf \: y = 2 \times 1 - 2 \\ [/tex]
[tex]\sf \: y = 2 - 2 \\ [/tex]
[tex]\implies\sf \: \boxed{\bf \: y=0 \: } \\ [/tex]
Hence,
[tex]\begin{gathered}\begin{gathered}\bf\: \implies\sf \: \begin{cases} &\sf{Number \: of \: pants \: purchased \: = \: 1} \\ \\ &\sf{Number \: of \: skirts \: purchased \: = \: 0} \end{cases}\end{gathered}\end{gathered}[/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional information
Let us consider pair of linear equations as
[tex]\sf \: a_1x + b_1y + c_1 = 0 \\ [/tex]
and
[tex]\sf \: a_2x + b_2y + c_2 = 0 \\ [/tex]
1. System of linear equations is consistent having unique solution iff
[tex]\qquad\boxed{ \sf{ \:\dfrac{a_1}{a_2} \:\ne \: \dfrac{b_1}{b_2} \: }} \\ [/tex]
2. System of linear equations is consistent having infinitely many solutions iff.
[tex]\qquad\boxed{ \sf{ \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \: }} \\ [/tex]
3. System of linear equations is inconsistent having no solution iff
[tex]\qquad\boxed{ \sf{ \:\dfrac{a_1}{a_2} \: = \: \dfrac{b_1}{b_2} \: \ne \: \dfrac{c_1}{c_2} \: }} \\ [/tex]
SOLUTION :-
As we know that,
The number of skirts is two less than twice the number of pants purchased = 2x - 2
the number of skirts is four less than four times
the number of pants she purchased = 4x - 4
on putting the value we get :-
2x - 2 = 4x - 4
4x - 2x = 4 - 2
2x = 2
x = 1
then,
on substituting the x = 1 we get :-
= 2(1) - 2
= 2 - 2
= 0
therefore,
number of pants purchased = 1
number of skirts purchased =0