Solve each problem. Molly bought 5.28 dollars worth of oranges and 8.80 dollars worth of apples.

step 1 We are going to do problem number 43 in this question. Molly buys two product that is apple and

Solve each problem. Molly bought 5.28 dollars worth of...

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more linear equations with the same set of variables. Solving such a system means finding the values of the variables that satisfy all equations simultaneously. This concept is fundamental in algebra and is widely applied in real-world problems where relationships between quantities need to be represented and resolved systematically.

Translating Word Problems into Equations

This concept involves converting a real-world problem described in words into algebraic expressions and equations. It requires identifying relationships, quantities, and conditions stated in the problem, and then forming equations that model these relationships. This translation is crucial as it bridges the gap between verbal information and mathematical analysis.

Variable Assignment

Variable assignment is the process of defining symbols to represent unknown quantities in a problem. By assigning variables to these unknowns, the problem solver can form equations that express the relationships between different elements. This abstraction simplifies complex relationships and is a critical step in algebraic problem-solving.

Substitution Method

The substitution method is a strategy used to solve systems of linear equations. It involves solving one equation for one variable and then substituting that expression into the other equation(s). This method reduces the number of variables in an equation, making it easier to solve the system step by step.

Unit Pricing and Rate Concepts

Unit pricing involves determining the cost per single unit of an item, which is essential for understanding and comparing different pricing scenarios. In problems concerning rates or ratios, establishing the unit rate allows for the formulation of equations that relate the total cost to the quantity purchased, a concept that is widely applicable in everyday life and commerce.

Transcript

00:01 We are going to do problem number 43 in this question.

00:04 Molly buys two product that is apple and orange.

00:09 Apple and orange.

00:13 The total amount paid by molly is $8 .80 for apple and $5 .28 for orange.

00:28 So this is the given information and what are the given information? and what are the more given information here so it is said that molly bought two more points of oranges than apple so let us say that molly bowers bought x pounds of apple so he will be buying x .2 pound of orange okay x plus 2 pound of orange and it is said that the cost of apple is twice that of orange so let us say that the cost of orange is y and if the cost of orange is y then the cost of apple will be twice means to y so this is the given information which is given to us so what we can write equation is we can write that the product of pound and cost the number of pound and cost it will give us the total cost of apple so you'll write here x into 2y this will give us 8 .8 zero dollars okay so here it comes to xy equals to 8 .80 dollars let's say this is equation number one now let us write equation number two this was this we had written for apple now we are going to write for orange now similarly in orange also the product of pound and cost will give us the total cost paid by molly so again we are going to multiply that that is x plus two into y this will give us 5 .28 okay so if you multiply this we will have xy plus 2y this is equals to 5 .28 dollars okay now what we are going to do is we are going to multiply lhs and rhs by 2 okay multiplying this equation both side by 2 so we will have here 2xy plus 4 y this will be equals to 10 .56 dollars okay so let us say this equation as equation number second okay now what we are going to do we are going to just subtract equation two with equation number one okay this is equation two and this is equation number one so we are going to operate this operation so if we are going to subtract it we will have 4 y equals to 10 .56 10 .56 minus 8 .80 okay so we'll have a 4y will be equals to 1 .76 so we got y that is equals to 0 .44 and what is why here why we had assumed the cost of each pound of each pound of orange okay so you got the we got the cost of each pound of orange so from here number let us write here total pounds of oranges total pounds of orange will be this is a total amount paid by molly was 5 .28 here this will be equals to 0 .44 so these two these two decimals will go out and we will just write here this will be equals to 12 pounds okay so we can say that molly bought 12 pounds of orange.

04:21 Now let's say how many pounds of apples...

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