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System (2): (2,- 6)
y_p=-1/2x+3
Explanation: See solution.
(II): Add (I)
(II): Remove parentheses
(II): Add and subtract terms
(II): .LHS /2.=.RHS /2.
Having solved for y, we can find x by substituting this value for y in the first equation.
The solution to (1) is (5,3). Let's also solve (2). Again, since both equations have y on the left-hand side and with opposite coefficients, we should use the Elimination Method.
(II): Add (I)
(II): Remove parentheses
(II): Add and subtract terms
(II): LHS+4x=RHS+4x
(II): .LHS /6.=.RHS /6.
We have found that x=2. By substituting this value for x in the second equation, we can find y.
The solution to (2) is (2,- 6).
c
y= mx+ b [0.2em]
[-1em]
&Slope: m &y-intercept: b
Since both graphs cut the y-axis at lattice points, we can identify the y-intercepts by examining the diagram.
Now we can start writing the equations. |c|c|c|c| Line &Equation & m & b q&y= mx+ 8 & m & 8 p&y= mx+ 3 & m & 3 We also have to find the lines slope. If we divide the vertical distance by the horizontal distance between two identifiable points on each graph, we can determine the slopes.
Now we can complete the equations. |c|c|c|c| Line &Equation & m & b [0.2em] [-0.8em] q&y= - 1/2x+ 8 & - 1/2 & 8 [0.8em] [-0.8em] p&y= 2x+ 3 & 2 & 3 [0.3em]
They are perpendicular.
y=- 12x+3 & (I) y=2x+8 & (II) Since both equations are solved for y, we should use the Substitution Method to solve the system.
(I): y= 2x+8
(I): LHS-3=RHS-3
(I): LHS * (- 2)=RHS* (- 2)
(I): LHS+4x=RHS+4x
(I): .LHS /5.=.RHS /5.
(I): Rearrange equation
The solution is (-2,4). To determine if it matches the solution in the graph, we have to find the x- and y-coordinate of the point where the lines intersect.
We are correct.