Answer:
Part 1) The solution is the point [tex](0.806,0.226)[/tex] (see the explanation)
Part 2) The solution is the point [tex](\frac{25}{31},\frac{7}{31})[/tex] (see the explanation)
Part 3) The solution is the point [tex](\frac{25}{31},\frac{7}{31})[/tex] (see the explanation)
Part 4) ordered pair (2,2)
Step-by-step explanation:
Part 1) Solve the system by graphing
we have
[tex]8x-2y=6[/tex] ----> equation A
[tex]3x+7y=4[/tex] ----> equation B
Solve the system by graphing
Remember that the solution of the system is the intersection point both lines
using a graphing tool
The solution is the point [tex](0.806,0.226)[/tex]
see the attached figure N 1
Part 2) Solve the system using the substitution method
we have
[tex]8x-2y=6[/tex] ----> equation A
[tex]3x+7y=4[/tex] ----> equation B
Isolate the variable y in the equation A
[tex]2y=8x-6[/tex]
[tex]y=4x-3[/tex] -----> equation C
Solve the system by substitution
substitute equation C in equation in equation B
[tex]3x+7(4x-3)=4[/tex]
solve for y
[tex]3x+28x-21=4[/tex]
[tex]31x=4+21[/tex]
[tex]x=\frac{25}{31}[/tex]
Find the value of y
[tex]y=4(\frac{25}{31})-3[/tex]
[tex]y=\frac{100}{31}-3[/tex]
[tex]y=\frac{7}{31}[/tex]
therefore
The solution is the point [tex](\frac{25}{31},\frac{7}{31})[/tex]
Part 3) Solve the system using the addition method
we have
[tex]8x-2y=6[/tex] ----> equation A
[tex]3x+7y=4[/tex] ----> equation B
Multiply equation A by 7 both sides
[tex]7(8x-2y)=7(6)[/tex]
[tex]56x-14y=42[/tex] -----> equation C
Multiply equation B by 2 both sides
[tex]2(3x+7y)=2(4)[/tex]
[tex]6x+14y=8[/tex] -----> equation D
Adds equation C and equation D
[tex]56x-14y=42\\6x+14y=8\\---------\\56x+6x=42+8\\62x=50\\x=\frac{50}{62}[/tex]
simplify
[tex]x=\frac{25}{31}[/tex]
Find the value of y
substitute the value of x in any equation
equation A
[tex]8(\frac{25}{31})-2y=6[/tex]
[tex]\frac{200}{31}-2y=6[/tex]
[tex]2y=\frac{200}{31}-6[/tex]
[tex]y=\frac{100}{31}-3[/tex]
[tex]y=\frac{7}{31}[/tex]
therefore
The solution is the point [tex](\frac{25}{31},\frac{7}{31})[/tex]
Part 4) Provide only one ordered pair (x, y) that will work for both system of inequalities. y>-2x+3 y<4x+1
we have
[tex]y>-2x+3[/tex] ----> inequality A
The solution of the inequality A is the shaded area above the dashed line
The equation of the dashed line A is [tex]y=-2x+3[/tex]
The slope of the dashed line is negative
The y-intercept of the dashed line is (0,3)
The x-intercept of the dashed line is (1.5,0)
[tex]y< 4x+1[/tex] ----> inequality B
The solution of the inequality B is the shaded area below the dashed line
The equation of the dashed line B is [tex]y=4x+1[/tex]
The slope of the dashed line is positive
The y-intercept of the dashed line is (0,1)
The x-intercept of the dashed line is (-0.25,0)
The solution of the system is the shaded area above the dashed line A and below the dashed line B
see the attached figure N 2
Remember that
If a ordered pair is a solution of the system of inequalities, then the ordered pair must lie in the shaded area region of the solution set (makes true both inequalities)
A solution of the system of inequalities is the ordered pair (2,2)
The point lie in the shaded area of the solution set
Substitute the value of x and the value of y of the ordered pair in each inequality
Verify inequality A
For x=2, y=2
[tex]2>-2(2)+3[/tex]
[tex]2>-1[/tex] ----> is true
so
The ordered pair satisfy inequality A (the ordered pair is a solution of the inequality A)
Verify inequality B
For x=2, y=2
[tex]2< 4(2)+1[/tex]
[tex]2< 9[/tex]----> is true
so
The ordered pair satisfy inequality B (the ordered pair is a solution of the inequality B)
The ordered pair is a solution of the system, because makes true both inequalities
