Ws Substitution

Solving systems of equations by substitution – 2009 By the end of this activity you will be able to solve systems of equations using substitution. 1. Warm up by solving the equation 3(4 − 3y) + 5y = −24

y = 310 − 51x y = 124 + 53x Notice the equations are already solved for y.

2. Procedure: In order to solve systems of equation with substitution, we first have to (1) solve one equation for one variable, then we (2) substitute the result into the other equation, (3) solve for the other variable, and finally, (4) substitute again and solve for remaining variable. Here’s an example 4x + 5y = 20

(1)

5x + 6y = 30

(2)

Solve for x in the first equation: 4x = 20 − 5y x = 5 − 1.25y

(3) (4)

Substitute the value of x, 5 − 1.25y into the other equation and solve for y. 5(5 − 1.25y) + 6y = 30

(5)

25 − 6.25y + 6y = 30

(6)

−.25y = 5 y = −20

(7) (8)

From here we can solve for x easily by substituting -20 back into the original equation. 4x + 5(−20) = 20

(9)

4x = 120

(10)

x = 30

(11)

So the answer is (30, −20). (a) Copy the example. (b) Try imitating this solution but instead, solve for y in the second equation first. 3. The train problem is a good problem to solve with substitution. Two trains leave Portland and Seattle at the same time on the same line. Portland is mile 124 on the Amtrak line and Seattle is mile 310. If the train to- wards Seattle travels at 53 miles per hour and the train towards Portland travels at about 51 mph, at around what mile mark will they cross?

(a) Solve the system of equations using substitution. 4. Solve the following system by substitution x + 4y = 153 y = 3x − 4 5. The digits in a two digit number have a sum of 15. When the digits are reversed the number is 9 less than the original. What is the original number? 6. There are 180 quarters and dimes in a jar. If the jar contains $33.30, give the number of quarters and dimes in the jar. 7. Solve the Tortoise and Hare Problem using substitution. The tortoise and the hare have devised a staggered race. The tortoise begins the race at mile mark 6 and sets off towards the end. At the same time, the hare sets off from mile mark 1. If the tortoise can cover five miles in four hours, and the hare can do seven miles in three hours, when will they cross paths and exactly how far will they have travelled?



Solutions

93 52 4743 y = 310 − 52 y ≈ 219



y = 310 − 51

1. Warm up by solving the equation 3(4 − 3y) + 5y = −24 12 − 9y + 5y = −24

The trains will cross close to the 219th mile marker.

−4y = −36 y=9 2. Procedure: In order to solve systems of equation with substitution, we first have to (1) solve one equation for one variable, then we (2) substitute the result into the other equation, (3) solve for the other variable, and finally, (4) substitute again and solve for remaining variable.

4. Solve the following system by substitution 13 + 4y = 153 4y = 140

x + 4y = 153

y = 35

y = 3x − 4 x + 4(3x − 4) = 153

Here’s an example 4x + 5y = 20

(12)

5x + 6y = 30

(13)

(a) Copy the example. (b) Try imitating this solution but instead, solve for y in the second equation first.   5 4x + 5 5 − x = 20 6 (18) 25 24 4x + 5y = 20 (14) x + 25 − x = 20 6 6 5x + 6y = 30 (15) (19) 6y = 30 − 5x x − = −5 (16) 6 (20) 5 y =5− x 6 x = 30 (17) (21) 3. The train problem is a good problem to solve with substitution. Two trains leave Portland and Seattle at the same time on the same line. Portland is mile 124 on the Amtrak line and Seattle is mile 310. If the train to- wards Seattle travels at 53 miles per hour and the train towards Portland travels at about 51 mph, at around what mile mark will they cross?

x + 12x − 16 = 153 13x = 169 x = 13

5. The digits in a two digit number have a sum of 15. When the digits are reversed the number is 9 less than the original. What is the original number? x + y = 15 9x − 9y = 9

10x + y − 9 = x + 10y

x = 15 − y 9(15 − y) − 9y = 9 135 − 18y = 9 −18y = −126 y=7 x + 7 = 15 x=8 So the original number was 10x + y or 87. 6. There are 180 quarters and dimes in a jar. If the jar contains $33.30, give the number of quarters and dimes in the jar. x + y = 180

y = 310 − 51x y = 124 + 53x

.25x + .1y = 33.3 x = 180 − y .25(180 − y) + .1y = 33.3

Notice the equations are already solved for y. (a) Solve the system of equations using substitution. 310 − 51x = 124 + 53x 186 = 104x 186 93 x= = ≈ 1.788461538 104 52

45 − .25y + .1y = 33.3 −.15y = −11.7 y = 78 x + 78 = 180 x = 102 There are 102 quarters and 78 dimes in the jar.

Solve the Tortoise and Hare Problem using substitution. The tortoise and the hare have devised a staggered race. The tortoise begins the race at mile mark 6 and sets off towards the end. At the same time, the hare sets off from mile mark 1. If the tortoise can cover five miles in four hours, and the hare can do seven miles in three hours, when will they cross paths and exactly how far will they have travelled?

5 x+6 4 7 y = x+1 3 5 7 x+6= x+1 4 3 5 7 x− x=5 3 4 y=

28x − 15x = 60

Tortoise

(22)

Hare

(23)

Substitution

(24)

Reflexive Prop + Addition Prop

(25)

· 60 Mult Prop

(26)

13x = 60

(27)

8 hr = 4.615384hr 13   5 60 y= +6 4 13

x=4

y =6+

10 75 mi = 11 mi = 11.769230mi 13 13

(28) Substitution

(29) (30)