zach invests $4000 part of it at 10% annual intrest and the rest at 12% annual intrest. if he earned $460 in intrest at the end of 1 year, how much did he invest at each rate. how do you solve that with system of equations?
+129840 I = Prt where I is the interest earned, P is the amount invested, and r is the interest rate.
Call the amount he invested at 10%, "x.' And call the amount that he invested at 12%, "4000-x." ...so we have
460 = .10x + .12(4000-x) simplify this
460 = .10x + 480 - .12x subtract 480 from both sides
-20 = -.02x
x = .(20)/(-.02) = 1000 and that's the amount invested at 10%
So, 4000-x = 4000-1000 = 3000...... and that's the amount invested at 12%
+129840 I = Prt where I is the interest earned, P is the amount invested, and r is the interest rate.
Call the amount he invested at 10%, "x.' And call the amount that he invested at 12%, "4000-x." ...so we have
460 = .10x + .12(4000-x) simplify this
460 = .10x + 480 - .12x subtract 480 from both sides
-20 = -.02x
x = .(20)/(-.02) = 1000 and that's the amount invested at 10%
So, 4000-x = 4000-1000 = 3000...... and that's the amount invested at 12%
