Question

In the following exercises, for each ordered pair, decide (a) is the ordered pair a solution to the equation? (b) is the point on the line? Graph can't copy $y=\frac{1}{2} x-3$; A: $(0,-3)$; B: $(2,-2)$; C: $(-2,-4)$; D: $(4,1)$ 

   In the following exercises, for each ordered pair, decide
(a) is the ordered pair a solution to the equation? (b) is the point on the line?
Graph can't copy
 $y=\frac{1}{2} x-3$;
A: $(0,-3)$; 
B: $(2,-2)$; 
C: $(-2,-4)$; 
D: $(4,1)$
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Intermediate Algebra
Intermediate Algebra
Lynn Marecek 1st Edition
Chapter 3, Problem 7 ↓
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In the following exercises, for each ordered pair, decide (a) is the ordered pair a solution to the equation? (b) is the point on the line? Graph can't copy $y=\frac{1}{2} x-3$; A: $(0,-3)$; B: $(2,-2)$; C: $(-2,-4)$; D: $(4,1)$
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Transcript

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00:01 In this problem, we are going to determine if each point is a solution to the equation.
00:06 So we're going to start out with the point zero negative three.
00:10 We're going to take our equation and fill in for y and x.
00:17 We're going to substitute.
00:18 So we're going to let x be zero and see if this left side of the equation equals negative three.
00:25 It's a question mark.
00:28 Okay.
00:28 So does negative 3 equals 0 minus 3? negative 3 does equal negative 3.
00:38 So 0 negative 3 is a solution to that equation...
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