Part d
in some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. let's say we now have
3c+4d=5
2c+5d=2
none of these terms has a coefficient of 1. instead, we'll pick the variable with the smallest coefficient and isolate it. move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient.
part e
now that you have one of the two variables in part d isolated, use substitution to solve for the two variables. you may want to review the multiplication and division of fractions and simplifying an expression primers. enter the answer as two numbers (either fraction or decimal), separated by a comma, with c first.

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge