Exercise 4 Page 241

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a In the first equation, the coefficient for

x

1 .

We only need to subtract

2 y

from the equation to isolate it. Then we can substitute the expression into the second equation.

{x + 2 y = - 7 2 x - y = - 9 (I) (II)

SubEqn

$(I):$ $LHS - 2 y = RHS - 2 y$

{x = - 7 - 2 y 2 x - y = - 9

Substitute

$(II):$ $x = - 7 - 2 y$

{x = - 7 - 2 y 2 (- 7 - 2 y) - y = - 9

▼

(II):

Solve for

y

Distr

$(II):$ Distribute $2$

{x = - 7 - 2 y - 14 - 4 y - y = - 9

AddEqn

$(II):$ $LHS + 14 = RHS + 14$

{x = - 7 - 2 y - 4 y - y = 5

SubTerm

$(II):$ Subtract term

{x = - 7 - 2 y - 5 y = 5

DivEqn

$(II):$ $LHS / (- 5) = RHS / (- 5)$

{x = - 7 - 2 y y = - 1

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b In the first equation, the coefficient for

x

1 .

We only need to add

2 y

to the equation to isolate it. Then we can substitute the expression into the second equation.

{x - 2 y = - 6 2 x + y = - 2 (I) (II)

AddEqn

$(I):$ $LHS + 2 y = RHS + 2 y$

{x = - 6 + 2 y 2 x + y = - 2

Substitute

$(II):$ $x = - 6 + 2 y$

{x = - 6 + 2 y 2 (- 6 + 2 y) + y = - 2

▼

(II):

Solve for

y

Distr

$(II):$ Distribute $2$

{x = - 6 + 2 y - 12 + 4 y + y = - 2

AddEqn

$(II):$ $LHS + 12 = RHS + 12$

{x = - 6 + 2 y 4 y + y = 10

AddTerms

$(II):$ Add terms

{x = - 6 + 2 y 5 y = 10

DivEqn

$(II):$ $LHS / 5 = RHS / 5$

{x = - 6 + 2 y y = 2

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c In the second equation, the coefficient for

y

1 .

We only need to add

2 x

to the equation to isolate it. Then we can substitute the expression into the first equation.

{- 3 x + 2 y = - 10 - 2 x + y = - 6 (I) (II)

AddEqn

$(II):$ $LHS + 2 x = RHS + 2 x$

{- 3 x + 2 y = - 10 y = - 6 + 2 x

Substitute

$(I):$ $y = - 6 + 2 x$

{- 3 x + 2 (- 6 + 2 x) = - 10 y = - 6 + 2 x

▼

(I):

Solve for

x

Distr

$(I):$ Distribute $2$

{- 3 x - 12 + 4 x = - 10 y = - 6 + 2 x

AddEqn

$(I):$ $LHS + 12 = RHS + 12$

{- 3 x + 4 x = 2 y = - 6 + 2 x

AddTerms

$(I):$ Add terms

{x = 2 y = - 6 + 2 x

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d In the second equation, the coefficient of

x

1 .

We only need to add

3 y

to the equation to isolate it. Then we can substitute the expression into the first equation.

{3 x + 2 y = 13 x - 3 y = - 3 (I) (II)

AddEqn

$(II):$ $LHS + 3 y = RHS + 3 y$

{3 x + 2 y = 13 x = - 3 + 3 y

Substitute

$(I):$ $x = - 3 + 3 y$

{3 (- 3 + 3 y) + 2 y = 13 x = - 3 + 3 y

▼

(I):

Solve for

y

Distr

$(I):$ Distribute $3$

{- 9 + 9 y + 2 y = 13 x = - 3 + 3 y

AddEqn

$(I):$ $LHS + 9 = RHS + 9$

{9 y + 2 y = 22 x = - 3 + 3 y

AddTerms

$(I):$ Add terms

{11 y = 22 x = - 3 + 3 y

DivEqn

$(I):$ $LHS / 11 = RHS / 11$

{y = 2 x = - 3 + 3 y

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e The coefficient of

x

in the second equation is

- 1 .

We can isolate it by changing the signs of the entire equation first. Then we can substitute it into the first equation.

{3 x - 2 y = 9 - x - 3 y = 8 (I) (II)

▼

(II):

Solve for

x

MultEqn

$(II):$ $LHS \cdot (- 1) = RHS \cdot (- 1)$

{3 x - 2 y = 9 x + 3 y = - 8

SubEqn

$(II):$ $LHS - 3 y = RHS - 3 y$

{3 x - 2 y = 9 x = - 8 - 3 y

Substitute

$(I):$ $x = - 8 - 3 y$

{3 (- 8 - 3 y) - 2 y = 9 x = - 8 - 3 y

▼

(I):

Solve for

y

Distr

$(I):$ Distribute $3$

{- 24 - 9 y - 2 y = 9 x = - 8 - 3 y

AddEqn

$(I):$ $LHS + 24 = RHS + 24$

{- 9 y - 2 y = 33 x = - 8 - 3 y

AddTerms

$(I):$ Add terms

{- 11 y = 33 x = - 8 - 3 y

DivEqn

$(I):$ $LHS / (- 11) = RHS / (- 11)$

{y = - 3 x = - 8 - 3 y

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f The coefficient of

y

in the first equation is

- 1 .

We can isolate it by changing the signs of the entire equation first. Then we can substitute it into the second equation.

{3 x - y = - 6 4 x + 5 y = 11 (I) (II)

▼

(I):

Solve for

y

MultEqn

$(I):$ $LHS \cdot (- 1) = RHS \cdot (- 1)$

{- 3 x + y = 6 4 x + 5 y = 11

AddEqn

$(I):$ $LHS + 3 x = RHS + 3 x$

{y = 6 + 3 x 4 x + 5 y = 11

Substitute

$(II):$ $y = 6 + 3 x$

{y = 6 + 3 x 4 x + 5 (6 + 3 x) = 11

▼

(II):

Solve for

x

Distr

$(II):$ Distribute $5$

{y = 6 + 3 x 4 x + 30 + 15 x = 11

SubEqn

$(II):$ $LHS - 30 = RHS - 30$

{y = 6 + 3 x 4 x + 15 x = - 19

AddTerms

$(II):$ Add terms

{y = 6 + 3 x 19 x = - 19

DivEqn

$(I):$ $LHS / 19 = RHS / 19$

{y = 6 + 3 x x = - 1

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Subchapter links

Communicate Your Answer p.241

Monitoring Progress p.242-244

Exercise 4 Page 241

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