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Pearson Algebra 1 Common Core, 2011 View details

6. Systems of Linear Inequalities

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Exercise 42 Page 405

a In order to find the value of

x,

we first have to find the slope of the line

2 x + y = 3 .

This will allow us to find the slope of a line perpendicular to the given one. The given equation is in the standard form, so we will rearrange it so that it is written in slope-intercept form first.

2 x + y = 3

SubEqn

$LHS - 2 x = RHS - 2 x$

y = 3 - 2 x

CommutativePropAdd

Commutative Property of Addition

y = - 2 x + 3

In this equation the coefficient of the

x

variable gives the slope of the line,

- 2 .

Perpendicular lines have negative reciprocals for their slopes. In this case, this will be

\frac{1}{2} .

Since we can calculate the slope using the two given points on the line,

(- 1, 6)

and

(x, 2),

we can use the Slope Formula to solve for the missing

x

value.

m = \frac{y _{1} - y _{2}}{x _{1} - x _{2}}

SubstituteValues

Substitute values

\frac{1}{2} = \frac{6 - 2}{- 1 - x}

▼

Solve for

x

SubTerm

Subtract term

\frac{1}{2} = \frac{4}{- 1 - x}

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

1 = \frac{8}{- 1 - x}

MultEqn

$LHS \cdot (- 1 - x) = RHS \cdot (- 1 - x)$

- 1 - x = 8

AddEqn

$LHS + 1 = RHS + 1$

- x = 9

MultEqn

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

x = - 9

b Now, having the slope and a point on the line, we can write a new equation using the point-slope form. We have found that the slope is

\frac{1}{2},

0.5,

and we can use the point

(- 1, 6),

m = 0.5

and

(x_{1}, y_{1}) = (- 1, 6) .

y - y_{1} = m (x - x_{1})

SubstituteValues

Substitute values

y - 6 = 0.5 (x - (- 1))

▼

Write in slope-intercept form

SubNeg

$a - (- b) = a + b$

y - 6 = 0.5 (x + 1)

Distr

Distribute $0.5$

y - 6 = 0.5 x + 0.5

AddEqn

$LHS + 6 = RHS + 6$

y = 0.5 x + 6.5

The equation of the line perpendicular to the given line

2 x + y = 3

and through the given points is

y = 0.5 x + 6.5 .

c To find a point that is a solution to both equations, we need to find their point of intersection. One way to do this is by equating the lines and solving for