The first equation is
(Equation 1)
The second equation is (Equation 2)
Putting the value of x from equation 1 in equation 2.
we get,
by simplifying the given equation,
Using discriminant formula,
Now the formula for solution 'x' of quadratic equation is given by:
Hence, these are the required solutions.
When you have this type of problem, you need to combine the like-terms and isolate the variable.
3x + 122 = 22x - 11
Add 11 to both sides to get rid of it
3x + 122 + 11 = 22x - 11 + 11 (-11 + 11=0)
3x + 133 = 22x
Then you would bring the 3x to the other side, so subtract 3x from both sides
3x + 133 = 22x
-3x -3x
133 = 22x - 3x
133 = 19x
Then divide both sides by 19 to isolate x
133/19 = 19x/19
133/19 = 7, so x = 7
Hope this helps!!
*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*
(21)
Area of a Regular Hexagon: square units
(22)
Similar to (21)
Area = square units
(23)
For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:
Hence, area of the hexagon will be: square units
(24)
Given is the inradius of an equilateral triangle.
Substituting the value of inradius and calculating the length of the side of the equilateral triangle:
Side = 16 units
Area of equilateral triangle = square units