Question 972177: 1.Lisa had twice as much money as Lin. When Lisa gave Lin 5c,both had the same amount. How much did Lisa have originally? Found 2 solutions by macston, solver91311:Answer by macston(5194) (Show Source):
You can put this solution on YOUR website! .
N=Lin amount; A=Lisa amount=2N
A-5=N+5 Substitute for A
2N-5=N+5 Add 5 to each side.
2N=N+10 Subtract N from each side.
N=10 Lin originally had 10c.
A=2N=2(10)=20
ANSWER: Lisa originally had 20c.
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R=cost for ride; G=amount for game
3R+4G=$11.50 Multipy by 4
12R+16G=$46.00 Call this Equation 1
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4R+3G=$13.00 Multiply by 3.
. 12R+9G=$39.00 Subtract equation 1 from this
.-12R+16G=$46.00
-7G=-$7.00 Divide each side by -7
G=$1.00 One game costs $1.00
3R+4G=$11.50 Put in value for G
3R+4($1.00)=$11.50
3R+$4.00=$11.50 Subtract $4.00 from each side.
3R=$7.50 Divide each side by 3
R=$2.50 One ride costs $2.50
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CHECK:
4R+3G=$13.00
4($2.50)+3($1.00)=$13.00
$10.00+$3.00=$13.00
$13.00=$13.00
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For one ride and one game:
ANSWER=R+G
ANSWER=$2.50+$1.00
ANSWER=$3.50
Let represent the amount that Lisa has originally. Let represent the amount that Lin has originally. Then is the amount that Lisa has after the transaction and is the amount that Lin has after the transaction.
Solve the 2X2 system for and . Hint: Solve by substitution.
John
My calculator said it, I believe it, that settles it
