Answer:
Yes, we can find a unique price for an apple and an orange.
Step-by-step explanation:
Let x be the price of one apple and y be the price of one orange.
We have been given that a fruit stand has to decide what to charge for their produce. They need $5.30 for 1 apple and 1 orange.
We can represent this information in an equation as:
[tex]x+y=5.30...(1)[/tex]
They also need $7.30 for 1 apple and 2 oranges.
[tex]x+2y=7.30...(2)[/tex]
Upon substituting our given information we formed a system of equations. Let us see if this system is solvable or not.
For a unique solution [tex]\frac{a_1}{a_2} \neq\frac{b_1}{b_2}[/tex], where [tex]a_1[/tex] and [tex]b_1[/tex] are constant of x and y variables of 1st equation respectively. [tex]a_2[/tex] and [tex]b_2[/tex] are constant of x and y variables of 2nd equation respectively.
Let us check our system of equations for unique solution.
[tex]\frac{1}{1} \neq\frac{1}{2}[/tex]
[tex]1 \neq\frac{1}{2}[/tex]
We can clearly see that 1 is not equal to half, therefore, we can find a unique price for an apple and orange using our system of equations.
Upon subtracting our 1st equation from 2nd equation we will get,
[tex]x-x+2y-y=7.30-5.30[/tex]
[tex]y=2[/tex]
Therefore, price of one orange is $2.
Upon substituting y=2 in equation 1 we will get,
[tex]x+2=5.30[/tex]
[tex]x=5.30-2[/tex]
[tex]x=3.30[/tex]
Therefore, price of one apple is $3.30.
Answer:
No; the system has no solution.
Step-by-step explanation:
A fruit stand has to decide what to charge for their produce. They decide to charge \$5.30$ 5, point, 30 for 111 apple and 111 orange. They also plan to charge \$14$14 dollar sign, 14 for 222 apples and 222 oranges. We put this information into a system of linear equations.
Can we find a unique price for an apple and an orange?
