Robert D. answered • 05/10/16

Tutor

5
(10)
College Physics Grad with Extensive Tutoring Experience

Let

**x**be the pounds of expensive candy.Let

**y**be the pounds of cheap candy.If there are

**10 more**pounds of cheap candy**y**than expensive candy**x,**then our first equation is

**y = x + 10**

which is convenient because we now have an expression for one of our variables

**y.**This equation expresses the quantity of candy we have

We can form another equation that expresses the monetary value of the candy we have.

Every pound of expensive candy

**x**costs**$3.50 .**Every pound of cheap candy

**y**costs**$1.00 .**We have

**$28**worth of candy all together.Therefore, our second equation is

**3.50x + 1.00y = 28.0**

So our system of equations is

**y = x + 10**

**3.50x + 1.00y = 28**

Plugging in our expression for

**y**into the second equation, we get**3.50x + 1.00(x + 10) = 28**

**3.50x + 1.00x +10 = 28**

**4.50x + 10 = 28**

**4.50x = 28 -10**

**x = 18/4.50**

**= 4**

We can plug this value for

**x**into the first equation and solve for**y****y = x + 10**

**y = (4) + 10**

**y = 14**

So we have

**x = 4**pounds of expensive candy and**y = 14**pounds of cheap candy.To double check, we can plug both values back into both equations, and if the statements are true, then the values are correct.

**y = x + 10**

**(14) = (4) + 10**

**14 = 14 TRUE!**

**3.50x + 1.00y = 28**

**3.50(4) + 1.00(14) = 28**

**14 + 14 = 28**

**28 = 28 TRUE!**