# SOLUTION: John invested a portion of \$15,000 at 10% annual interest and the rest at 6% annual interest. If he earned \$1,260 for the year from the two accounts, how much did he invest at 10%?

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: John invested a portion of \$15,000 at 10% annual interest and the rest at 6% annual interest. If he earned \$1,260 for the year from the two accounts, how much did he invest at 10%?      Log On

 Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Word Problems: Mixtures Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Mixture Word Problems Question 537454: John invested a portion of \$15,000 at 10% annual interest and the rest at 6% annual interest. If he earned \$1,260 for the year from the two accounts, how much did he invest at 10%?Found 2 solutions by mananth, lmeeks54:Answer by mananth(12270)   (Show Source): You can put this solution on YOUR website!Investment I 6.00% per annum ---x Investment II 10.00% per annum ---y x + y= 15000 ------------------------1 6.00% x + 10.00% y= = \$1,260.00 Multiply by 100 6 x + 10 y= = \$126,000.00 --------2 Multiply (1) by -6 we get -6 x -6 y= = -90000.00 Add this to (2) 0 x 4 y= = \$36,000.00 divide by 4 y = \$9,000.00 investment at 10.00% Balance \$6,000.00 investment at 6.00% CHECK \$6,000.00 --------- 6.00% ------- \$360.00 \$9,000.00 ------- 10.00% ------- \$900.00 Total -------------- \$1,260.00 m.ananth@hotmail.ca Answer by lmeeks54(105)   (Show Source): You can put this solution on YOUR website!Let H = # \$ invested at 10% Let L = # \$ invested at 6% ... Given: H + L = 15,000 .10H + .06L = 1,260 ... With a system of two equations with two unknowns, rewrite one of the equations so that it is written for one unknown in terms of the other unknown. ... H + L = 15,000 H = 15,000 - L ... substitute H = 15,000 - L into the 2nd equation: .10(15,000 - L) + .06L = 1,260 ... simplify by combining like terms: 1,500 - .10L + .06L = 1,260 ... subtract 1,500 from both sides: -.04L = -240 ... divide both sides by -.04: L = \$6,000 ... This represents the amount of money invested at 6%. To find the amount of money invested at 10%, go back to our H = 15,000 - L equality and solve for H: ... H = 15,000 - 6,000 H = \$9,000 ... H = \$9,000 equals the amount of money invested at 10%. ... Check your work by inserting H = 9,000 and L = 6,000 into the 2nd equation: .10(9,000) + .06(6,000) = 1,260 900 + 360 = 1,260 1,260 = 1,260 checks... ... Cheers, Lee