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Question: How do you solve\[2f(5f - 2) - 10({f^2} - 3f + 6) = - 8f(f + 4) + 4(2{f^2} - 7f)\]?...

How do you solve2f(5f−2)−10(f2−3f+6)=−8f(f+4)+4(2f2−7f)2f(5f - 2) - 10({f^2} - 3f + 6) = - 8f(f + 4) + 4(2{f^2} - 7f)?

Explanation

Solution

We solve left hand and right hand side of the equation separately first. Multiply the terms outside the bracket with all terms inside the bracket and solve by adding or subtracting terms with like variables on both sides of the equation. Then bring all variables and constant values to one side of the equation making the right hand side of the equation 0. Cancel possible factors and write the value of the variable i.e. f as simplest form of fraction.

Complete step by step answer:
We have to solve 2f(5f−2)−10(f2−3f+6)=−8f(f+4)+4(2f2−7f)2f(5f - 2) - 10({f^2} - 3f + 6) = - 8f(f + 4) + 4(2{f^2} - 7f)
We solve the left hand side and right hand side of the equation simultaneously. Open the brackets on each side and multiply terms outside the bracket with terms inside the bracket
⇒(2f×5f)−(2f×2)−(10×f2)+(10×3f)−(10×6)=−(8f×f)−(8f×4)+(4×2f2)−(4×7f)\Rightarrow (2f \times 5f) - (2f \times 2) - (10 \times {f^2}) + (10 \times 3f) - (10 \times 6) = - (8f \times f) - (8f \times 4) + (4 \times 2{f^2}) - (4 \times 7f)Calculate each of the products in the bracket
⇒10f2−4f−10f2+30f−60=−8f2−32f+8f2−28f\Rightarrow 10{f^2} - 4f - 10{f^2} + 30f - 60 = - 8{f^2} - 32f + 8{f^2} - 28f
Combine terms having same variable on both sides of the equation
⇒(10f2−10f2)+(30f−4f)−60=(−8f2+8f2)−(32f+28f)\Rightarrow \left( {10{f^2} - 10{f^2}} \right) + \left( {30f - 4f} \right) - 60 = \left( { - 8{f^2} + 8{f^2}} \right) - \left( {32f + 28f} \right)
Solve each bracket on both sides of the equation
⇒0+26f−60=0−60f\Rightarrow 0 + 26f - 60 = 0 - 60f
Bring all values to left hand side of the equation
⇒26f−60+60f=0\Rightarrow 26f - 60 + 60f = 0
Combine terms with same variable on left hand side of the equation
⇒(26f+60f)−60=0\Rightarrow \left( {26f + 60f} \right) - 60 = 0
Solve the bracket on left hand side of the equation
⇒86f−60=0\Rightarrow 86f - 60 = 0
Shift constant value to right and side of the equation
⇒86f=60\Rightarrow 86f = 60
Cancel same factors on both sides of the equation
⇒43f=30\Rightarrow 43f = 30
Divide both sides of the equation by 43
⇒43f43=3043\Rightarrow \dfrac{{43f}}{{43}} = \dfrac{{30}}{{43}}
Cancel same factors from numerator and denominator on left hand side of the equation
⇒f=3043\Rightarrow f = \dfrac{{30}}{{43}}

The solution of 2f(5f−2)−10(f2−3f+6)=−8f(f+4)+4(2f2−7f)2f(5f - 2) - 10({f^2} - 3f + 6) = - 8f(f + 4) + 4(2{f^2} - 7f) is f=3043f = \dfrac{{30}}{{43}}

Note: Students are likely to make mistakes while shifting the values from one side of the equation to another side of the equation as they forget to change the sign of the value shifted. Keep in mind we always change the sign of the value from positive to negative and vice versa when shifting values from one side of the equation to another side of the equation. Also, students see the quadratic equation and they start solving using a determinant method, always checking first if the terms cancel out or not and then apply formula.