Lesson 5-4

Example 1 Multiply One Equation to Eliminate
	Use elimination to solve the system of equations.
	x + 3y = -4
	x + 2y = 9
Multiply the first equation by –3 so the coefficients of the x-terms are additive inverses. Then add the equations.
	x + 3y = -4	Multiply by –3.	 -x – 9y = 12
	x + 2y = 9	(+) x + 2y =  9
		-7y = 21			Add the equations.
		   = 	Divide each side by –7.
		y = -3
	Now substitute –3 for y in either equation to find the value of x.
		 x + 2y = 9	Second equation
		x + 2(-3) = 9	y = -3
		   x – 6 = 9	Simplify.
			  x = 15	Add 6 to each side and simplify.
	The solution is (15, -3).

Example 2 Multiply Both Equations to Eliminate
	Use elimination to solve the system of equations.
	5x – 7y = -2
	-4x + 6y = 4
	Method 1	Eliminate x.
	5x – 7y = -2	Multiply by 4.	 20x – 28y = -8
	-4x + 6y = 4		Multiply by 5.	(+) -20x + 30y = 20
				   2y = 12		Add the equations.
				     = 	Divide each side by 2.
				      y = 6		Simplify.
	Now substitute 6 for y in either equation to find the value of x.
		5x – 7y = -2	First equation
		5x – 7(6) = -2	y = 6
		5x – 42 = -2	Simplify.
		  5x – 42 + 42 = -2 + 42	Add 42 to each side.
			  5x = 40	Simplify.	
			   x = 8	Divide each side by 5 and simplify.
	The solution is (8, 6).

	Method 2	Eliminate y.
	5x – 7y = -2	Multiply by 6.	 30x – 42y = -12
	-4x + 6y = 4		Multiply by 7.	(+) -28x + 42y =   28
				  2x 	 =   16		Add the equations.
					 = 		Divide each side by 2.
					  x = 8				Simplify.
	Now substitute 8 for x in either equation to find the value of y.
		5x – 7y = -2		First equation
		5(8) – 7y = -2		x = 8
		40 – 7y = -2		Simplify.
		  40 – 7y – 40 = -2 – 40		Subtract 40 from each side.
			    -7y = -42		Simplify.
			    = 		Divide each side by –7.
			     y = 6		Simplify.
	The solution is (8, 6), which matches the result obtained in Method 1.

Standardized Test EXAMPLE 	
Example 3 						Writing Systems of Equations
FIELD TRIP  Joan is taking her pre-school students and parent helpers to see a play at the high school. There will be twenty people attending the play. Tickets cost $2 for children and $4 for adults. The total cost to attend the play is $50. Which system of equations can be used to determine how many adult helpers will be attending thee play?
A      a + c = 50		B    a + c = 20	              C     a + c = 20				D      a - c = 20	
     4a + 2c = 20		   4a + 2c = 50							  2a + 4c = 50					  4a +2 c = 50		

Read the Test Item
You are asked to find a system of equations to represent this situation using a, the number of adult helpers and c, the number of children attending the play.

Solve the Test Item
Represent the situation algebraically by writing two equations.

The total number of children and adults is 20.
		a + c = 20
One equation is a + c = 20.

The number of adult tickets times $2 each plus the number of children tickets times $2 each is $50.
4a + 2c = 50
The second equation is 4a + 2c = 50.

The system of equations that represents this situation is a + c = 20 and 4a + 2c = 50. The answer is B.

Example 4 Write and Solve a System of Equations
A jet travels 2600 miles with the wind in 4 hours. The return trip into the wind took 5 hours. Find the rate of the jet with no wind.
Variables 		Let j = the rate of the jet with no wind and w = the rate of the wind. Use the formula rate  time = distance, or rt = d.

Equations
	r	t	d	rt = d	
With the Wind	j + w	4	2600	4j + 4w = 2600	
Into the Wind	j – w	5	2600	5j – 5w = 2600	

Use elimination with multiplication to solve this system. Since the problem asks for j, eliminate w.

4j + 4w = 2600	Multiply by 5.	20j + 20w = 13000
5j – 5w = 2600	Multiply by 4.	(+) 20j – 20w = 10400
	40j	= 23400			Add the equations.
	 = 		Divide each side by 40.
	j = 585					Simplify.
The rate of the jet in no wind is 585 miles per hour.