Name: Duane Who is asking: Other Level of the question: All Question: Hey i am pretty lost with this problem. Pat invested a total of \$3000 dollars. part of the money yields 10 percent interest per year and the rest yields 8 percent interest per year if the total yearly interest is \$256 how much did pat invest at 10 per cent and how much at 8 percent. I am not sure of the formula to solve this equation i though it was to divide out 10 and 8 percent out of 3000 then subtract. Hi Duane, You have the correct idea. You need to divide the \$3000 into 2 parts and apply 10% to one part and 8% to the other. If you only knew one part you could find the other since the sum is \$3000. This is a place where the notation of algebra really helps. Suppose that \$x was invested at 10% them \$(3000 - x) was invested at 8%. In one year, \$x invested at 10% would yield an interest of 0.10 x. In one year, \$(3000 - x) invested at 8% would yield an interest of 0.08 (3000 - x). But you know that the total interest is \$256 and hence 0.10 x + 0.08 (3000 - x) = 256 Solve for x. Penny