Exercise 1 Page 644

Practice makes perfect

a Our friend claims that any system formed by three of the following equations has exactly one solution.

Equation
$3 x + y + 3 z = 6$
$x + y + z = 2$
$4 x - 2 y + 4 z = 8$
$x - y + z = 2$
$2 x + y + z = 4$
$3 x + y + 9 z = 12$

We want to find a system of three equations that supports his claim. For example, we can pick the following three equations.

Equation
$3 x + y + 3 z = 6$
$x + y + z = 2$
$4 x - 2 y + 4 z = 8$
$x - y + z = 2$
$2 x + y + z = 4$
$3 x + y + 9 z = 12$

To check whether a system of the three highlighted equations has only one solution, we need to solve it.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + y + z = 2 x - y + z = 2 2 x + y + z = 4 (I) (II) (III)

To do this we will use the Elimination Method. Notice that the coefficients of the variables

y

and

z

in Equation (I) are the same as in Equation (III).

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + y + z = 2 x - y + z = 2 2 x + y + z = 4 (I) (II) (III)

Let's subtract Equation (III) from Equation (I) to eliminate variables

y

and

z .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + y + z = 2 x - y + z = 2 2 x + y + z = 4

SysEqnSub

$(I):$ Subtract $(III)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + y + z - (2 x + y + z) = 2 - 4 x - y + z = 2 2 x + y + z = 4

▼

(I):

Simplify

Distr

$(I):$ Distribute $(- 1)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + y + z - 2 x - y - z = 2 - 4 x - y + z = 2 2 x + y + z = 4

SubTerms

$(I):$ Subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - x = - 2 x - y + z = 2 2 x + y + z = 4

MultEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 2 x - y + z = 2 2 x + y + z = 4

We found that

x = 2 .

Let's substitute that for

x

in Equations (II) and (III).

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 2 x - y + z = 2 2 x + y + z = 4 (I) (II) (III) ⇕ ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 2 2 - y + z = 2 2 (2) + y + z = 4 (I) (II) (III)

Equations (II) and (III) form a system of two equations with two variables.

{2 - y + z = 2 2 (2) + y + z = 4

At this point we can solve it as we would with any other systems of two equations. First, let's simplify them.

{2 - y + z = 2 2 (2) + y + z = 4

▼

Simplify

SubEqn

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

{- y + z = 0 2 (2) + y + z = 4

Multiply

$(II):$ Multiply

{- y + z = 0 4 + y + z = 4

SubEqn

$(II):$ $LHS - 4 = RHS - 4$

{- y + z = 0 y + z = 0

SysEqnAdd

$(I):$ Add $(II)$

{- y + z + y + z = 0 y + z = 0

AddTerms

$(I):$ Add terms

{2 z = 0 y + z = 0

DivEqn

$(I):$ $LHS / 2 = RHS / 2$

{z = 0 y + z = 0

We found that

z = 0 .

To find the value of

y,

we can substitute

z = 0

into the second equation.

{z = 0 y + z = 0

Substitute

$(II):$ $z = 0$

{z = 0 y + 0 = 0

IdPropAdd

$(II):$ Identity Property of Addition

{z = 0 y = 0

We solved the system of equations and found only one solution:

x = 2,

y = 0,

and

z = 0 .

As we were solving the system we did not stumble upon any identities like

0 = 0

or contradictions such as

1 = 0 .

Therefore, this system of equations has only one solution and it supports our friends claim.

b This time, we want to show that our friend's claim is incorrect. To do that, we want to find a system that does not have only one solution.

Equation
$3 x + y + 3 z = 6$
$x + y + z = 2$
$4 x - 2 y + 4 z = 8$
$x - y + z = 2$
$2 x + y + z = 4$
$3 x + y + 9 z = 12$

For example, we can pick the following three equations.

Equation
$3 x + y + 3 z = 6$
$x + y + z = 2$
$4 x - 2 y + 4 z = 8$
$x - y + z = 2$
$2 x + y + z = 4$
$3 x + y + 9 z = 12$

To check whether a system of the three highlighted equations has only one solution, we need to solve it.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + y + 3 z = 6 x + y + z = 2 x - y + z = 2 (I) (II) (III)

To do this we will use the Elimination Method. Notice that the coefficients of the variables

x

and

z

in Equation (II) are the same as in Equation (III).

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + y + 3 z = 6 x + y + z = 2 x - y + z = 2 (I) (II) (III)

Let's subtract Equation (III) from Equation (II) to eliminate variables

x

and

z .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + y + 3 z = 6 x + y + z = 2 x - y + z = 2

SysEqnSub

$(II):$ Subtract $(III)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + y + 3 z = 6 x + y + z - (x - y + z) = 2 - 2 x - y + z = 2

▼

(II):

Simplify

Distr

$(II):$ Distribute $(- 1)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + y + 3 z = 6 x + y + z - x + y - z = 2 - 2 x - y + z = 2

SubTerms

$(II):$ Subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + y + 3 z = 6 2 y = 0 x - y + z = 2

DivEqn

$(II):$ $LHS / 2 = RHS / 2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + y + 3 z = 6 y = 0 x - y + z = 2

We found that

y = 0 .

Let's substitute for

y

in Equations (I) and (III).

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + y + 3 z = 6 y = 0 x - y + z = 2 ⇕ ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + 0 + 3 z = 6 y = 0 x - 0 + z = 2

Equations (I) and (III) form a system of two equations with two variables.

{3 x + 0 + 3 z = 6 x - 0 + z = 2

At this point we can solve it as we would do with any other systems of two equations. First, let's simplify them.

{3 x + 0 + 3 z = 6 x - 0 + z = 2

▼

Simplify

$(I):$ & $(II):$ Add and subtract terms

{3 x + 3 z = 6 x + z = 2

DivEqn

$(I):$ $LHS / 3 = RHS / 3$

{x + z = 2 x + z = 2

SysEqnSub

$(I):$ Subtract $(II)$

{x + z - (x + z) = 2 - 2 x + z = 2

Distr

$(I):$ Distribute $(- 1)$

{x + z - x - z = 2 - 2 x + z = 2

DivEqn

$(I):$ $LHS / 2 = RHS / 2$

{0 = 0 x + z = 2

We obtained an identity,

0 = 0 .

This means that this system has infinitely many solutions. Let's try to describe them. We know that

y = 0

and that

x + z = 2 .

{y = 0 x + z = 2 ⇕ {y = 0 z = 2 - x

Therefore, any ordered triple

(x, y, z)

of the form

(x, 0, 2 - x)

is a solution to the system to the three selected equations. This means that our friend's claim is incorrect.

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Exercises p.644-645

Exercise 1 Page 644

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