Question

Solve the following equation: \[17(x + 4) + 8(x + 6) = 11(x + 5) + 15(x + 3)\]

Answer 1

Hint: We solve the given equation for the value of unknown variable i.e. ‘x’. Multiply the values outside the bracket to the terms inside the bracket using distributive law of multiplication over addition. Write equations on both sides in simplest form. Shift all variable values on one side and constant values on the other side of the equation. Calculate the value of the variable by dividing with suitable value.
* Distributive Property: For any three numbers ‘a’, ‘b’ and ‘c’ we can write \[a(b + c) = ab + bc\]

Complete step-by-step solution:
We are given the equation \[17(x + 4) + 8(x + 6) = 11(x + 5) + 15(x + 3)\] … (1)
Use distributive property on each of the brackets in left hand side and right hand side of the equation (1)
\[ \Rightarrow (17 \times x + 17 \times 4) + (8 \times x + 8 \times 6) = (11 \times x + 11 \times 5) + (15 \times x + 15 \times 3)\]
Calculate each of the products inside each of the brackets in left hand side and right hand side of the equation
\[ \Rightarrow (17x + 68) + (8x + 48) = (11x + 55) + (15x + 45)\]
Open all brackets on both sides of the equation
\[ \Rightarrow 17x + 68 + 8x + 48 = 11x + 55 + 15x + 45\]
Add the terms with variables together and terms as constants together on each side of the equation
\[ \Rightarrow 25x + 116 = 27x + 100\]
Bring all constants on one side of the equation and all terms with variable on other side of the variable
\[ \Rightarrow 116 - 100 = 27x - 25x\]
Calculate the differences on both sides of the equation
\[ \Rightarrow 16 = 2x\]
Divide both sides of the equation by 2
\[ \Rightarrow \dfrac{{16}}{2} = \dfrac{{2x}}{2}\]
Cancel same factors from numerator and denominator on both sides of the equation i.e. 2
\[ \Rightarrow 8 = x\]
So, the value of x is 8

\[\therefore \]The solution of the equation \[17(x + 4) + 8(x + 6) = 11(x + 5) + 15(x + 3)\] is \[x = 8\]

Note: Many students make mistake of solving the equation by bringing all the terms on one side of the equation i.e. by dividing the terms in LHS by terms in RHS which is wrong, keep in mind we would have divided the terms in LHS by terms in RHS if we were given the RHS as single term, RHS is given as sum of two values and it is difficult to proceed in this manner.
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.