# A man has a total of 90 pesos. He has six more 5-peso coins than 1-peso coins. How many coins of...

## Question:

A man has a total of 90 pesos. He has six more 5-peso coins than 1-peso coins. How many coins of each type does the man have?

## Algebraic equation word problems

An algebraic equation is formed when a relation between any variable is equated to a constant.

In the word problems of algebraic equation, this relation is given in statement and we need to form an algebraic equation out of it. Assuming a base variable and determining other unknowns in its term also helps in simplifying the equation.

For example: 4x +3y = 1 is an algebraic equation where x and y are variables and 1 is the equivalent constant of the relation shown in the equation.

Let the number of {eq}5\ pesos {/eq} and {eq}1\ peso {/eq} coins are {eq}x {/eq} and {eq}y. {/eq}

According to the problem: The man has six more 5-peso coins than 1-peso coins.

$$x= y+ 6$$

Total value is {eq}90\ pesos. {/eq}

$$5x + y = 90$$

Substituting value of {eq}x=y+6 {/eq}

$$5(y +6) + y = 90$$

$$5y +30 + y = 90$$

$$6y = 90 -30$$

$$6y = 60$$

$$y = 10$$

Using the value of {eq}y {/eq} in the equation {eq}x=y + 6 {/eq}

$$x= y+ 6$$

$$x= 10+ 6$$

$$x= 16$$

The number of {eq}5\ pesos {/eq} and {eq}1\ peso {/eq} coins are {eq}16 {/eq} and {eq}10. {/eq} 