Consider the following problem:<\/p>\n
Sue has $2200 to invest for one year. \u00a0She has a choice two investments: \u00a0investment A pays 4%, while investment B pays 5.4%. \u00a0She wants to earn $100 interest for the year. How much should she invest in each choice?<\/p>\n
Solution:<\/p>\n
Let x=amount of money she puts in A (at 4%)<\/p>\n
Let y=amount of money she puts in B (at 5.4%)<\/p>\n
Now, we know that<\/p>\n
x + y = 2200 (equation 1)<\/p>\n
To get the percent of a number, you change the percent into a decimal (by dividing by 100), then multiply. \u00a0For instance, 29% of $246=0.29($246)=$71.34.<\/p>\n
Therefore, 4% of x is 0.04x and 5.4% of y is 0.054y. \u00a0We have<\/p>\n
0.04x + 0.054y = 100 (equation 2)<\/p>\n
We can solve this system of two equations using substitution: \u00a0notice that equation 1 implies<\/p>\n
y = 2200 – x<\/p>\n
We can substitute 2200 – x, instead of y, in equation 2:<\/p>\n
0.04x + 0.054(2200 – x) = 100<\/p>\n
Now we simplify and solve:<\/p>\n
0.04x + 118.8 – 0.054x = 100<\/p>\n
Subtract 118.8 from both sides:<\/p>\n
0.04x – 0.054x = -18.8<\/p>\n
Simplify the left:<\/p>\n
-0.014x = -18.8<\/p>\n
Divide both sides by -0.014:<\/p>\n
x = -18.8\/-0.014 = 1342.86<\/p>\n
Now, looking back at equation 1<\/p>\n
x + y = 2200<\/p>\n
we see that<\/p>\n
1342.86 + y = 2200<\/p>\n
Subtracting 1342.86 from both sides gives<\/p>\n
y = 857.14<\/p>\n
Looking back at our let statements, we see that<\/p>\n
x = amount invested at 4%<\/p>\n
y = amount invested at 5.4%<\/p>\n
We now know that she should invest $1342.86 at 4% and $857.14 at 5.4%.<\/p>\n
Check: \u00a01342.86(0.04) + 857.14(0.054) = 53.71 + 46.29 = 100<\/p>\n
From a high school math point of view, this question is settled.<\/p>\n
Questions arise about the investment aspect of this problem. \u00a0However, they belong to a different discipline. \u00a0We might explore some of those questions in future posts:)<\/p>\n