Question

Gail bought 5 pounds of oranges ans 2 pounds of bananas for $14. Her husband later bought 3 pounds of oranges ans 6 pounds of bananas for $18. What was the cost per pounds of the oranges and the bananas?

Answer 1

Answer:

Step-by-step explanation:

5r + 2b = 14 (5 lbs of oranges and 2 lbs of bananas = $14)

3r + 6b = 18 (3 lbs of oranges and 6 lbs of bananas = $18)

Let's use the elimination method. Multiply the first equation by -3:

-15r - 6b = -42

3r + 6b = 18

-----------------

12r = 24

Divide by 12:

r = 2

Answer 2

Answer:

The price per pound of the oranges is $ 2.

The price per pound of the bananas is $ 2.

Step-by-step explanation:

A system of linear equations is a set of equations that have more than one unknown. The unknowns appear in several of the equations, but not necessarily in all. In these equations the unknowns are related to each other.

In this case a system of equations is applied. For that, you must first define the unknowns (or also called variable):

• x: price per pounds of the orange
• y: price per pounds of the banana

The price of the 5 pounds of oranges, whose price is "x", and 2 pounds of bananas, whose price is "y", bought by Gail must add $ 14. Then expressed in an equation this represents: 5*x+2*y=14 Equation (A)

The price of the 3 pounds of oranges, whose price is "x", and 6 pounds of bananas, whose price is "y", bought by her husband must add $ 18. Then expressed in an equation this represents: 3*x+6*y=18 Equation (B)

Equation A and equation B form the equation system, where the variables are "x" and "y", which are the price of an orange and a banana:

$\left \{ {{5*x+2*y=14} \atop {3*x+6*y=18}} \right.$

One of the ways to solve a system of equations is through the substitution method.  This method consists of isolating one of the unknowns (for example, "x") and replacing its expression in the other equation. In this way, a first degree equation is obtained with the other variable, "y".

In this case, x is isolated from the first equation, obtaining:

5*x+2*y=14

5*x=14-2*y

$x=\frac{14-2*y}{5}$

$x=\frac{14}{5} -\frac{2}{5}*y$ Equation (C)

Replacing this equation (C) in the other equation you get:

$3*[\frac{14}{5} -\frac{2}{5} *y]+6*y=18$

Isolating the value of "y" you get:

$3*\frac{14}{5} -3*\frac{2}{5} *y+6*y=18$

$\frac{42}{5} -\frac{6}{5} *y+6*y=18$

$-\frac{6}{5} *y+6*y=18-\frac{42}{5}$

$\frac{24}{5} *y=\frac{48}{5}$

$y=\frac{\frac{48}{5} }{\frac{24}{5} }$

$y=\frac{48}{5} *\frac{5}{24}$

y=2

Remembering that "y" is the price per pounds of the banana, then it is possible to say that the price per pound of the banana is $ 2.

Replacing the value of the variable "y" in equation (c) and isolating the value of the variable "x" gives:

$x=\frac{14}{5} -\frac{2}{5}*2$

$x=\frac{14}{5} -\frac{4}{5}$

$x=\frac{14-4}{5}=\frac{10}{5}$

x=2

Remembering that "x" is the price per pounds of the oranges, then it is possible to say that the price per pound of the oranges is $ 2.