that would yield 770.420 + 770 = 1190So it works.ANSWERShe invested 7000 at 6%She invested 11000 at 7%-The first equation is the amount in acct. 1 plus the amount in acct 2 to give you the totalx+y=18000the next is the amt in each acct times its interest rate.0.07x+0.......
Add the two equations together.
y = 11000
Now you simply substitute 11000 for y in the first equation
x + (11000) = 18000
x = 7000
To check the answers, try running through the question. If she invested 7000 at 6%, that would yield 420. If she invested 11000 at 7%, that would yield 770.
420 + 770 = 1190
So it works.
ANSWER
She invested 7000 at 6%
She invested 11000 at 7%
The first equation is the amount in acct. 1 plus the amount in acct 2 to give you the total
x+y=18000
the next is the amt in each acct times it's interest rate.
0.07x+0.06y=1190
solve the first equation for y to get y=18000-x and plug that into the second equation
0.07x+0.06(18000-x)=1190, then solve for x
0.07x+1080-0.06x=1190
0.01x=110
x=11000, then plug that into the first equation to get the amount in the other acct
11000+y=18000.....y=7000
okay so... you have x and y. x is the amount of money in the first account. y is the second.
we know that x+y=18,000
we also know that .06x+.07y=1,190
okay, we can solve it. we have two variables and two equations.
take the first eq. move y to the other side (subtract y from both sides)
x=18000-y
plug that into the second equation...
.06(18000-y)+.07y=1190
distribute .06 into both terms in the parentheses. you get: 1080-.06y+.07y=1190
collect like terms:.01y=110
solve for y: y=11000
so... then you plug your newly found y into the first equation
x+11000=18000
x=7000
So... you find that the bank with 7% had $11000 in it. and the bank with 6% interest had $7000 in it.
