Respuesta :
Answer:
b. (-1, 1.5)
d. (0,3)
Step-by-step explanation:
we have
[tex]4y-6x=12[/tex]
we know that
If a ordered pair is on the graph of the linear equation, the the ordered pair must satisfy the linear equation
Verify each case
case a) we have
(-4,3)
substitute the value of x and the value of y in the equation
[tex]4(3)-6(-4)=12[/tex]
[tex]36=12[/tex] ----> is not true
The ordered pair not satisfy the equation
therefore
The ordered pair is not on the graph of the linear equation
case b) we have
(-1,1.5)
substitute the value of x and the value of y in the equation
[tex]4(1.5)-6(-1)=12[/tex]
[tex]12=12[/tex] ----> is true
The ordered pair satisfy the equation
therefore
The ordered pair is on the graph of the linear equation
case c) we have
(0,-2)
substitute the value of x and the value of y in the equation
[tex]4(-2)-6(0)=12[/tex]
[tex]-8=12[/tex] ----> is not true
The ordered pair not satisfy the equation
therefore
The ordered pair is not on the graph of the linear equation
case d) we have
(0,3)
substitute the value of x and the value of y in the equation
[tex]4(3)-6(0)=12[/tex]
[tex]12=12[/tex] ----> is true
The ordered pair satisfy the equation
therefore
The ordered pair is on the graph of the linear equation
case e) we have
(3,4)
substitute the value of x and the value of y in the equation
[tex]4(4)-6(3)=12[/tex]
[tex]-6=12[/tex] ----> is not true
The ordered pair not satisfy the equation
therefore
The ordered pair is not on the graph of the linear equation
case f) we have
(6,4)
substitute the value of x and the value of y in the equation
[tex]4(4)-6(6)=12[/tex]
[tex]-12=12[/tex] ----> is not true
The ordered pair not satisfy the equation
therefore
The ordered pair is not on the graph of the linear equation
The only points that a valid points on the graph of 4y - 6x = 12 are;
Option A; (-4,-3)
Option B; (-1, 1.5)
Option D; (0,3)
We are given the equation;
4y - 6x = 12
Let us make y the subject to get;
4y = 12 + 6x
y = (12 + 6x)/4
y = 3 + ³/₂x
Option A; at (-4, -3), let us put -4 for x to see if we will get y as -3;
y = 3 + ³/₂(-4)
y = 3 - 6
y = -3
Thus, (-4, -3) is a valid point
Option B; at (-1, 1.5)
y = 3 + ³/₂(-1)
y = 3 - 1.5
y = 1.5
Thus, (-1, 1.5) is a valid point
Option C; at (0, -2)
y = 3 + ³/₂(0)
y = 3
Thus, (0, -2) is not a valid point
Option D; at (0, 3)
y = 3 + ³/₂(0)
y = 3
Thus, (0, 3) is a valid point
Option E; At (3, 4)
y = 3 + ³/₂(3)
y = 3 + 4.5
y = 7.5
Thus; (3, 4) is not a valid point
Option F; At (6, 4)
y = 3 + ³/₂(6)
y = 3 + 9
y = 12
Thus, (6, 4) is not a valid point
In conclusion the only valid points are; (-4, -3), (-1, 1.5), (0, 3)
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