Zahra wants the equation below to have an infinite number of solutions when the missing number is placed in the box. _(x-3)+2x=-
2 answers:
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Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by
The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same
Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by
The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same
Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
See attached photo.
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Consider the equation:
Subtracting '4' from both the sides of the equation, we get as
Squaring on both the sides of the equation, we get
Subtracting '2' from both the sides of the equation, we get
x=14
Since, An extraneous solution is a solution that arises from the solving process that is not really a solution at all. But, in this equation x=14 is the solution of the given equation.
Hence, it is not an extraneous solution.