Exercise 1 Page 457

We are asked to set up and solve a system of linear equations to find how many nickels and dimes are in the jar. To do so, we must know that the value of a nickel is $ 0.05 and the value of a dime is $ 0.10. Let's write the verbal phrases as algebraic phrases.

Verbal phrase	Algebraic phrase
The number of nickels plus the number of dimes is 30	n + d = 30
The number of nickels plus the number of dimes have a total value of $ 2.10	0.05n + 0.10d = 2.10

Therefore, the system of equations is as follows. n+d=30 & (I) 0.05n+0.10d=2.10 & (II) Note that neither variable has equal or opposite coefficients in the equations. This means we first need to multiply one or both equations by some value to create equal or opposite coefficients. Then we can solve using the Elimination Method. n+d=30 0.05n+0.10d=2.10 (II)* 10 n+d=30 0.5n+d=21 To solve using the Elimination Method, we will start by subtracting Equation (II) from Equation (I).

n+d=30 & (I) 0.5n+d=21 & (II)

SysEqnSub

(I): Subtract (II)

n+d-( 0.5n+d)=30- 21 0.5n+d=21

▼

Solve Equation (I) for n

Distr

(I): Distribute - 1

n+d-0.5n-d=30-21 0.5n+d=21

AddSubTerms

(I): Add and subtract terms

0.5n=9 0.5n+d=21

DivEqn

(I): .LHS /0.5.=.RHS /0.5.

n=18 0.5n+d=21

We have found that n=18. To find the value of d, we will substitute 18 for n in Equation (II) and solve for d.

n=18 0.5n+d=21

Substitute

(II): n= 18

n=18 0.5( 18)+d=21

Multiply

(II): Multiply

n=18 9+d=21

SubEqn

(II): LHS-9=RHS-9

n=18 d=12

We have found that d=12. Therefore, the solution of the system of equations, which is the point of intersection of the lines, is (12,18). In the context of the problem, it means that there are 12 dimes and 18 nickels in the jar.