QUESTION IMAGE
Question
find the x- and y-intercepts. (lesson 5-2)
- $6x - 2y = 12$ 46. $-3y + x = 15$ 47. $4y - 40 = -5x$
tell whether each ordered pair is a solution of the given system. (lesson 6-1)
- $(3, 0)$; $\
$ 49. $(-1, 4)$; $\
$ 50. $(5, 6)$; $\
$
s of equations and inequalities
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Problems 45-47: Find x- and y-intercepts
For x-intercept: Set $y=0$, solve for $x$.
For y-intercept: Set $x=0$, solve for $y$.
Problem 45: $6x - 2y = 12$
Step1: Find x-intercept ($y=0$)
$6x - 2(0) = 12$
$6x = 12$
$x = 2$
Step2: Find y-intercept ($x=0$)
$6(0) - 2y = 12$
$-2y = 12$
$y = -6$
Problem 46: $-3y + x = 15$
Step1: Find x-intercept ($y=0$)
$-3(0) + x = 15$
$x = 15$
Step2: Find y-intercept ($x=0$)
$-3y + 0 = 15$
$-3y = 15$
$y = -5$
Problem 47: $4y - 40 = -5x$ (rewrite as $5x + 4y = 40$)
Step1: Find x-intercept ($y=0$)
$5x + 4(0) = 40$
$5x = 40$
$x = 8$
Step2: Find y-intercept ($x=0$)
$5(0) + 4y = 40$
$4y = 40$
$y = 10$
Substitute $x,y$ from the pair into both equations. If both are true, it is a solution.
Problem 48: $(3,0)$;
Step1: Substitute into first equation
$2(3) - 0 = 6
eq -6$ (False)
Step2: Substitute into second equation
$3 + 0 = 3$ (True)
Since one equation is false, the pair is not a solution.
Problem 49: $(-1,4)$;
Step1: Substitute into first equation
$4 - 2(-1) = 4 + 2 = 6$ (True)
Step2: Substitute into second equation
$-1 + 4(4) = -1 + 16 = 15$ (True)
Both equations are true, so the pair is a solution.
Problem 50: $(5,6)$;
Step1: Substitute into first equation
$\frac{1}{3}(6) + 5 = 2 + 5 = 7$ (True)
Step2: Substitute into second equation
$2(5) = 10
eq 12$ (False)
Since one equation is false, the pair is not a solution.
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(45-47):
- x-intercept: $(2, 0)$; y-intercept: $(0, -6)$
- x-intercept: $(15, 0)$; y-intercept: $(0, -5)$
- x-intercept: $(8, 0)$; y-intercept: $(0, 10)$
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