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### Linear Equations; Rates of Pay

```
Date: 12/16/97 at 12:29:20
From: michael
Subject: Linear equations

1) 3x-y+z = 0                   2) 2x^2+5y^2-53 = 0  3)x^2+2xy+x = 18
x+2y-z = -1                     4x^2+3y^2-43 = 0         2x-y = 5
1/(2x)+2(3Y)-1(3z) = -1/(3)

4) y^2+3xy+3y = -8    5) x^2+3xy-y^2 = -3
3y^2+xy+y = 8         x^2-xy-y^2 = 1

A carpenter and his assistant working together for 8 hours earn a
total of \$96. One day, the carpenter worked alone for 5 hours and was
joined by his assistant for the other 3 hours. On that day, they
earned a total of \$81. What are their rates of pay?
```

```
Date: 12/16/97 at 16:27:57
From: Doctor Rob
Subject: Re: Linear equations

1) Solve the first equation for z in terms of x and y. Substitute for
z in the other two equations. Solve the new second equation for y
in terms of x. Substitute for y in the new third equation. This
will give you one equation in x alone. Solve it for x, and use that
solution to figure out y, and use both to determine z.

2) Solve the first equation for y^2 in terms of x^2. Substitute for
y^2 in the second equation. This will give you one equation in x^2.
Solve it for x^2, and use that solution to figure out y^2. Take
square roots to get x and y, and don't forget that either can be
positive or negative. There will be four pairs (x,y) of solutions.

3) Start with the second equation, since it has lower degree. Solve it
for y in terms of x, and substitute in the first equation. This
gives you one quadratic equation in x. Solve it for x, getting two
roots. Use each of these to figure out the corresponding values of
y. There will be two pairs (x,y) of solutions.

4) Solve the first equation for x in terms of y. Substitute the
result in the second equation. Expand everything in sight, clear
fractions, and bring all terms over to one side of the equation.
This is one quadratic equation in one unknown y. Find the roots
of this equation. Use these values to find the values of x
corresponding to them.  There will be two pairs (x,y) of solutions.

5) You could proceed as above, but there is a cleverer way. Subtract
one equation from the other. This will tell you what x*y equals.
Solve that equation for y in terms of x. Substitute that back into
either of the original equations. Clear fractions. Bring all terms
over to one side. Solve for x. Use that value to figure out y.
There will be two pairs (x,y) of solutions.

Let C be the carpenter's hourly wage, and A the assistant's hourly
wage. 8*(C + A) = 96, and 5*C + 3*(C+A) = 81. Use the methods above to
solve.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Linear Algebra

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