Answer:
<u><em>(-1,1)</em></u>
Step-by-step explanation:
We can solve this by either graphing and finding ther point the lines intersect, or mathematically, I'll do both.
<u>Graphing:</u>
<u>Mathematically:</u>
−2x + 4y = 6
y = 2x + 3
See the attached graph. The lines intersect at (-1,1)
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I'll rearrange the first equation (to make it easier for me):
−2x + 4y = 6
4y = 2x + 6
y = (1/2)x + 1.5
Now lets substitute the second equation into the first so that we can eliminate y:
y = 2x + 3
[(1/2)x + 1.5] = 2x + 3
- (3/2)x = (3/2)
x = -1
If x = -1:
y = 2(-1) + 3
y = 1
The solution is x = -1 and y = 1, or (-1,1)
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Both approaches give us (-1,1), the solution to the system of equations. It is the only point that satisfies both equations.
Answer:
(x, y) = (2, 9)
Step-by-step explanation:
For the triangles to be congruent, the hypotenuses must be the same length:
y = x + 7
and the marked leg must be the same length in each triangle:
y -3 = 4x -2
These are two equations in two unknowns (a "system" of equations) that can be solved in any of the usual ways. Since the first equation gives an expression for y, it is convenient to substitute that into the second equation:
(x +7) -3 = 4x -2
x +4 = 4x -2 . . . . . . collect terms
x +6 = 4x . . . . . . . . .add 2
6 = 3x . . . . . . . . . . . subtract x
2 = x . . . . . . . . . . . . divide by 3
y = 2 + 7 = 9 . . . . . .substitute for x in the first equation
The values you're looking for are x = 2, y = 9.
Answer:
E
Step-by-step explanation:
i guess the dotted lines outline a square
so get the area of the square which is 6×6=36
then don't focus on the shaded part but unshaded you'll see two right angled triangles
you will get a total for both as 21
then get the area of the square 36-21=15
so the area becomes 15
Answer:
Subtract from both sides of the equation the term you don't want
Step-by-step explanation:
In solving equations, you generally want to "undo" operations that are done to the variable. Addition is "undone" by adding the opposite (that is, subtracting the amount that was added). Multiplication is "undone" by division.
If you have variables on both sides of the equation, pick one of the variable terms and subtract it from both sides of the equation.
<u>Example</u>
2x = x +1
If we choose to subtract x, then we will have a variable term on the left and a constant term on the right:
2x -x = x -x +1 . . . . . . . x is subtracted from both sides
x = 1 . . . . . . simplify
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Note that we purposely set up this example so that removing the variable term from the right side caused the variable term and constant term to be on opposite sides of the equal sign. It may not always be that way. As long as you remember that an unwanted term can be removed by subtracting it (from both sides of the equation), you can deal with constant terms and variable terms no matter where they appear.
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<em>Additional Comment</em>
It usually works well to choose the variable term with the smallest (or most negative) coefficient. That way, when you subtract it, you will be left with a variable term that has a positive coefficient.
