Ok, here is the problem "Bob has 60 coins consisting of quarters and dimes. the coins combined value is $9.45. Find out how many quarters and dimes there are."
Ps- We are using substitution etc to solve this, and showing your work would be appreciated, thanks
+128448 OK...we can actually do it without substitution, but.....anyway
Let x be the number of dimes and y be the number of quarters
So
x + y = 60 or y = 60- x
And the number of dimes times their value plus the number of quarters times their value = $9.45 ....or...in equation form...
.10x + .25y = 9.45 and substituting for y, we have
.10x + .25(60 - x) = 9.45 multiply by 100 to clear the decimals
10x + 25(60-x) = 945 simplify
10x + 1500 - 25x = 945
-15x + 1500 = 945 subtract 1500 from both sides
-15x = -555 divide both sides by -15
x = 37 and that's the number of dimes
And (60 - 37) = 23 and that's the number of quarters
+128448 OK...we can actually do it without substitution, but.....anyway
Let x be the number of dimes and y be the number of quarters
So
x + y = 60 or y = 60- x
And the number of dimes times their value plus the number of quarters times their value = $9.45 ....or...in equation form...
.10x + .25y = 9.45 and substituting for y, we have
.10x + .25(60 - x) = 9.45 multiply by 100 to clear the decimals
10x + 25(60-x) = 945 simplify
10x + 1500 - 25x = 945
-15x + 1500 = 945 subtract 1500 from both sides
-15x = -555 divide both sides by -15
x = 37 and that's the number of dimes
And (60 - 37) = 23 and that's the number of quarters
