Answer:
1) C [tex]s=\left \{ (-2,-9) \right \}[/tex] 2) (-2,0) 3)[tex]S=\left \{ (14,10) \right \}[/tex] 4) [tex]S=\left \{ \left ( 2,\frac{1}{2} \right ) \right \}[/tex] 5) c )S={(0,3)} 6) (0,2) 7) [tex]A) S=\left \{ \left ( 1,4 \right ) \right \}[/tex] 8) Missing graph 9) Vertical line (check below) 10) B 11)
Step-by-step explanation:
1) Solving by the Addition/Elimination Method. Firstly, let's reduce one variable by making some algebraic adjustments and then adding it up:
[tex]\left\{\begin{matrix}-2x&+y&=5 *(3)\\ 3x&-2y&=12 *(2)\end{matrix}\right.\Rightarrow \left\{\begin{matrix}-6x&+3y&=15 \\6x&-4y&=24 \end{matrix}\right.\Rightarrow -y=9\Rightarrow y=-9\Rightarrow 6x-4(-9)=24\Rightarrow 6x+36=24\Rightarrow 6x=-36+24\Rightarrow 6x=-12\Rightarrow x=-2\Rightarrow s=\left \{ -2,-9 \right \}[/tex]
2) Solving by Substitution Method. Where y=x+2 is plugged in the 2nd equation.
[tex]\left\{\begin{matrix}y =&x+2 & \\ 2x-y &=-4 & \end{matrix}\right.\Rightarrow 2x-(x+2)=-4\Rightarrow 2x-x-2=-4\Rightarrow x=-4+2\Rightarrow x=-2\: and\: y=-2+2\Rightarrow y=0\Rightarrow S=\left \{ \left ( -2,0 \right ) \right \}[/tex]
3) Solving it, again, by the Substitution Method due to the I equation form:
[tex]\left\{\begin{matrix}x=y+4& \\ 2x-3y=-2& \end{matrix}\right.\Rightarrow 2(y+4)-3y=-2\Rightarrow 2y+8-3y=-2\Rightarrow y=10\: \: x=10+4\Rightarrow x=14\Rightarrow S=\left \{ (14,10) \right \}[/tex]
4) By the Addition Method
[tex]\left\{\begin{matrix}x+2y &=3 \\ 3x-2y &=5 \end{matrix}\right.\Rightarrow 4x=8\Rightarrow x=2\Rightarrow 2+2y=3\Rightarrow 2y=1\Rightarrow y=\frac{1}{2}\: S=\left \{ \left ( 2,\frac{1}{2} \right ) \right \}[/tex]
5) To use the graph method to solve the system of Linear equations is possible by graphing each equation on the Cartesian Plane.
Check the graph below, this system has only one solution.
c)S={(0,3)}
6) Solving y=-1/3x+2 y=x+2
(Check the graph below)
A) A) (0, 2)
7) Solving by Substitution Method:
[tex]\left\{\begin{matrix}x=y-3 & \\ x+3y=13& \end{matrix}\right.\Rightarrow y-3+3y=13\Rightarrow 4y=13+3\Rightarrow 4y=16\Rightarrow y=4\: \: \\x=4-3\Rightarrow x=1 \\S=\left \{ \left ( 1,4 \right ) \right \}[/tex]
8) Missing graph
9)
[tex]\frac{1}{2}x+x=1+2\Rightarrow \frac{3}{2}x=3\Rightarrow x=2\\ S=\{2}\\[/tex]
Check the graph below its answer
10) Solving by the Addition Method
[tex]\left\{\begin{matrix}x+3y=5 & \\ -x+6y=4& \end{matrix}\right.\Rightarrow 9y=9\Rightarrow y=1 \: \: And \: \: x+3(1)=5\Rightarrow x=5-3\Rightarrow x=2\\S=\left \{ \left ( 2,1 \right ) \right \}[/tex]
11) Sorry, missing graph for the question.
12) Sorry, missing graph for the question.
13) D Check the graph below
14) Sorry, missing graph for the question.
A system of linear equations can be solved graphically, by substitution, by elimination, and by the use of matrices.
(1) -2x + y = -5 and 3x - 2y = 12
Make y the subject in -2x + y = -5
[tex]\mathbf{y = 2x - 5}[/tex]
Substitute [tex]\mathbf{y = 2x - 5}[/tex] in [tex]\mathbf{3x - 2y = 12}[/tex]
[tex]\mathbf{3x - 2(2x - 5) = 12}[/tex]
[tex]\mathbf{3x - 4x + 10 = 12}[/tex]
[tex]\mathbf{ - x + 10 = 12}[/tex]
Subtract 10 from both sides
[tex]\mathbf{ - x = 2}[/tex]
[tex]\mathbf{ x = -2}[/tex]
Recall that: [tex]\mathbf{y = 2x - 5}[/tex]
[tex]\mathbf{y = 2 \times -2 -5}[/tex]
[tex]\mathbf{y = -9}[/tex]
The solution is (-2,-9)
(2) y = x + 2 and 2x - y = -4
Substitute [tex]\mathbf{y = x + 2}[/tex] in [tex]\mathbf{2x - y = -4}[/tex]
[tex]\mathbf{2x - x - 2 = -4}[/tex]
[tex]\mathbf{x - 2 = -4}[/tex]
Add 2 to both sides
[tex]\mathbf{ x = -2}[/tex]
Recall that: [tex]\mathbf{y = x + 2}[/tex]
[tex]\mathbf{y =-2+ 2}[/tex]
[tex]\mathbf{y = 0}[/tex]
The solution is (-2,-0)
(3) x = y + 4 and 2x - 3y = -2
Substitute [tex]\mathbf{x = y + 4}[/tex] in [tex]\mathbf{2x - 3y = -2}[/tex]
[tex]\mathbf{2(y+4) - 3y = -2}[/tex]
[tex]\mathbf{2y+8 - 3y = -2}[/tex]
Collect like terms
[tex]\mathbf{2y- 3y = -2-8}[/tex]
[tex]\mathbf{-y = -10}[/tex]
[tex]\mathbf{y = 10}[/tex]
Recall that: [tex]\mathbf{x = y + 4}[/tex]
[tex]\mathbf{x = 10 + 4 = 14}[/tex]
The solution is (14,10)
(4) x + 2y = 3 and 3x - 2y = 5
Make x the subject in x + 2y = 3
[tex]\mathbf{x = 3 - 2y}[/tex]
Substitute [tex]\mathbf{x = 3 - 2y}[/tex] in [tex]\mathbf{3x - 2y = 5}[/tex]
[tex]\mathbf{3(3 - 2y) - 2y = 5}[/tex]
[tex]\mathbf{9 - 6y - 2y = 5}[/tex]
[tex]\mathbf{9 - 8y = 5}[/tex]
Collect like terms
[tex]\mathbf{- 8y = 5 - 9}[/tex]
[tex]\mathbf{- 8y = - 4}[/tex]
Divide by -4
[tex]\mathbf{y = \frac 12}[/tex]
Recall that: [tex]\mathbf{x = 3 - 2y}[/tex]
[tex]\mathbf{x = 3 - 2 \times 1/2 = 2}[/tex]
The solution is (2,1/2)
(5) 2x + y = 3 and x + y = 3
See attachment for the graphs of 2x + y = 3 and x + y = 3
From the attachment, the solution is: (0,3)
(6) y = -1/3x + 2 and y = x + 2
See attachment for the graphs of y = -1/3x + 2 and y = x + 2
From the attachment, the solution is: (0,2)
(7) x = y - 3 and x + 3y = 13
See attachment for the graphs of x = y - 3 and x + 3y = 13
From the attachment, the solution is: (1,4)
(8, 11, 12, 14) Incomplete
(9) See attachment for graph
(10) x + 3y = 5 and -x + 6y = 4
See attachment for the graphs of x + 3y = 5 and -x + 6y = 4
From the attachment, the solution is: (b) x = 2, y = 1
(13) y - x = -2 and 2x + y = 7
See attachment for the graphs of y - x = -2 and 2x + y = 7
From the attachment, the solution is: (3,1)
Read more about systems of linear equations at:
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