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Click right here to check out ALL problems on FinanceConcern 175658: Please help me settle this equation: Pat invested a total of $3,000. Part of the money returns 10 percent interestper year, and also the remainder yields 8 percent interest per year. If the complete yearlyinteremainder from this investment is $256, how a lot did Pat invest at 10 percentand also how much at 8 percent? Found 2 options by jim_thompson5910, Mathtut:Answer by jim_thompson5910(35256) (Sexactly how Source): You deserve to put this solution on YOUR website! Let x=amount invested at 10% and also y=amount invested at 8%"Pat invested a full of $3,000" equates to "Part of the money returns 10 percent interestper year, and the rest yields 8 percent interemainder per year. If the total yearlyinterest from this investment is $256" translates to Multiply both sides by 100 to obtain So we have the mechanism of equations:Let"s deal with by substitutionNow in order to solve this mechanism by making use of substitution, we have to fix (or isolate) one variable. I"m going to settle for y.So let"s isolate y in the first equation Start through the first equation Subtract from both sides Rearrange the equation---------------------Due to the fact that , we deserve to now relocation each in the second equation with to settle for Plug in right into the second equation. In other words, relocation each with . Notice we"ve eliminated the variables. So we currently have an easy equation with one unrecognized. Distribute to Multiply Combine like terms on the left sideSubtract 24000 from both sides Combine favor terms on the best side Divide both sides by 2 to isolate x Divide-----------------First Answer------------------------------So the initially part of our answer is: Due to the fact that we understand that we deserve to plug it right into the equation (remember we formerly resolved for in the initially equation).

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Start with the equation wbelow was formerly isolated. Plug in Multiply Combine prefer terms -----------------2nd Answer------------------------------So the second component of our answer is: -----------------Summary------------------------------So our answers are: and also So this indicates that Pat invested $800 at 10% and $2,200 at 8% Answer by Mathtut(3670) (Sexactly how Source): You have the right to put this solution on YOUR website! let x and y be the quantities invested at 10% and also 8%:x+y=3000........eq 1.1x+.08y=256....eq 2:recompose eq 1 to x=3000-y and also plug that worth right into eq2:.1(3000-y)+.08y=256:300-.1y+.08y=256:-.02y=-44:$amount invested at 8%:$amount invested at 10%