Respuesta :
Data:
A (apple)
B (banana)
Lydia → 5A + 3B = 8.50 (I)
Ari → 3A + 2B = 5.25 (II)
Solving: System of equations that represents the given the situation.
[tex] \left \{ {{5A + 3B = 8.50\:(I)} \atop {3A + 2B = 5.25\:(II) }} \right. [/tex]
Now let's find the values "A" and "B", we simulated to find the value of one of the unknowns:
[tex]\left \{ {{5A + 3B = 8.50\:*(-2)} \atop {3A + 2B = 5.25\:*(3) }} \right.[/tex]
[tex]\left \{ {{-10A -\diagup\!\!\!\!6B = -17.00} \atop {9A + \diagup\!\!\!\!6B = 15.75}} \right.[/tex]
[tex]-A=-1.25\:simplify\:*(-1)[/tex]
[tex]\boxed{\boxed{A=\$\:1.25}}\end{array}}\qquad\quad\checkmark[/tex]
Now substitute one of the equations to find the value of "B":
[tex]5A + 3B = 8.50\:(I)[/tex]
[tex]5*(1.25) + 3B = 8.50[/tex]
[tex]6.25 +3B = 8.50[/tex]
[tex]3B = 8.50-6.25[/tex]
[tex]3B = 2.25[/tex]
[tex]B = \frac{2.25}{3} [/tex]
[tex]\boxed{\boxed{B = \$\:0.75}}\end{array}}\qquad\quad\checkmark[/tex]
Answer:
Therefore, a system of equations that represents the given the situation.
[tex]\boxed{\boxed{\left \{ {{5A + 3B = 8.50\:(I)} \atop {3A + 2B = 5.25\:(II) }} \right.}}[/tex]
A (apple)
B (banana)
Lydia → 5A + 3B = 8.50 (I)
Ari → 3A + 2B = 5.25 (II)
Solving: System of equations that represents the given the situation.
[tex] \left \{ {{5A + 3B = 8.50\:(I)} \atop {3A + 2B = 5.25\:(II) }} \right. [/tex]
Now let's find the values "A" and "B", we simulated to find the value of one of the unknowns:
[tex]\left \{ {{5A + 3B = 8.50\:*(-2)} \atop {3A + 2B = 5.25\:*(3) }} \right.[/tex]
[tex]\left \{ {{-10A -\diagup\!\!\!\!6B = -17.00} \atop {9A + \diagup\!\!\!\!6B = 15.75}} \right.[/tex]
[tex]-A=-1.25\:simplify\:*(-1)[/tex]
[tex]\boxed{\boxed{A=\$\:1.25}}\end{array}}\qquad\quad\checkmark[/tex]
Now substitute one of the equations to find the value of "B":
[tex]5A + 3B = 8.50\:(I)[/tex]
[tex]5*(1.25) + 3B = 8.50[/tex]
[tex]6.25 +3B = 8.50[/tex]
[tex]3B = 8.50-6.25[/tex]
[tex]3B = 2.25[/tex]
[tex]B = \frac{2.25}{3} [/tex]
[tex]\boxed{\boxed{B = \$\:0.75}}\end{array}}\qquad\quad\checkmark[/tex]
Answer:
Therefore, a system of equations that represents the given the situation.
[tex]\boxed{\boxed{\left \{ {{5A + 3B = 8.50\:(I)} \atop {3A + 2B = 5.25\:(II) }} \right.}}[/tex]
Price for :
apple : $1.25
banana : $0.75
Further Explanation
One variable linear equation is an equation that has a variable and the exponent number is one.
Can be stated in the form:
[tex] \large {\boxed {\bold {ax = b}} [/tex]
or
ax + b = c, where a, b, and c are constants, x is a variable
Whereas the two-variable linear equation is a linear equation that has 2 variables and the exponent is one
Can be stated in the form:
[tex] \large {\boxed {\bold {ax + bx = c}}} [/tex]
x, y = variable
In this equation system there are coefficients and variables whose solutions can use:
1. substitution
2. elimination
3. graph
4. mix
Steps for completion with the substitution method:
1. change one of the equations to the form y = ax + b or x = ay + b
2. substitution the value of x / y to another equation
3. solve the equation
4. substitution results from point 3 in one equation
Lydia buys 5 pounds of apples and 3 pounds of bananas for a total of $ 8.50
We express it in algebraic form
5A + 3B = 8.5 equation 1
Ari buys 3 pounds of apples and 2 pounds of bananas for a total of $ 5.25
We express it in algebraic form
3A + 2B = 5.25 equation 2
From equation 1 and 2, e use elimination and substitution
1. elimination
5A + 3B = 8.5 x 2
3A + 2B = 5.25 x 3
10A + 6B = 17
9A + 6B = 15.75
------------------------
A = 1.25
2. substitution
5A + 3B = 8.5 equation 1
5(1.25) + 3B = 8.5
B = 0.75
Learn more
the substitution method
https://brainly.com/question/11705106
https://brainly.com/question/7162056
The work of a student to solve a set of equations
https://brainly.com/question/10589573