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Basic Math Examples
1.1814=(1y)⋅((0.83333)(1+0.2⋅y2))1.75
Step 1
Step 1.1
Simplify terms.
Step 1.1.1
Apply the distributive property.
1.1814=1y⋅(0.83333⋅1+0.83333(0.2y2))1.75
Step 1.1.2
Multiply.
Step 1.1.2.1
Multiply 0.83333 by 1.
1.1814=1y⋅(0.83333+0.83333(0.2y2))1.75
Step 1.1.2.2
Multiply 0.2 by 0.83333.
1.1814=1y⋅(0.83333+0.166666y2)1.75
1.1814=1y⋅(0.83333+0.166666y2)1.75
Step 1.1.3
Combine 1y and (0.83333+0.166666y2)1.75.
1.1814=(0.83333+0.166666y2)1.75y
1.1814=(0.83333+0.166666y2)1.75y
Step 1.2
Simplify the numerator.
Step 1.2.1
Factor 0.166666 out of 0.83333+0.166666y2.
Step 1.2.1.1
Factor 0.166666 out of 0.83333.
1.1814=(0.166666⋅5+0.166666y2)1.75y
Step 1.2.1.2
Factor 0.166666 out of 0.166666⋅5+0.166666y2.
1.1814=(0.166666(5+y2))1.75y
1.1814=(0.166666(5+y2))1.75y
Step 1.2.2
Apply the product rule to 0.166666(5+y2).
1.1814=0.1666661.75(5+y2)1.75y
Step 1.2.3
Raise 0.166666 to the power of 1.75.
1.1814=0.04347426(5+y2)1.75y
1.1814=0.04347426(5+y2)1.75y
1.1814=0.04347426(5+y2)1.75y
Step 2
Graph each side of the equation. The solution is the x-value of the point of intersection.
y≈0.73685674,2.45446785
Step 3