To solve the equation y = x^2 - 3x - 20 using systems, you can set up a system of equations by substituting y with x^2 - 3x - 20. This gives you the equation x^2 - 3x - 20 = 0. You can then solve this quadratic equation using factoring, the quadratic formula, or completing the square to find the values of x. Once you have the values of x, you can substitute them back into the original equation to find the corresponding values of y.
Solve using the quadratic formula
You cant solve it unless it is an equation. To be an equation it must have an equals sign.
Solve, using the Rule of 72 rate = 4%, years = 18, fv=$8,000. Solve for PV. Formula: PV = $1/(1+r) t PV = $8000/(1+.04) 18 PV = $8000/2.0258 3949.03 = $8000/2.20258
(2,-2)
Since the second equation is already solved for "y", you can replace "y" by "9" in the other equation. Then solve the new equation for "x".
Solve using the quadratic formula
8840-026
You cant solve it unless it is an equation. To be an equation it must have an equals sign.
You can solve lineaar quadratic systems by either the elimination or the substitution methods. You can also solve them using the comparison method. Which method works best depends on which method the person solving them is comfortable with.
Solve, using the Rule of 72 rate = 4%, years = 18, fv=$8,000. Solve for PV. Formula: PV = $1/(1+r) t PV = $8000/(1+.04) 18 PV = $8000/2.0258 3949.03 = $8000/2.20258
the answer
(2,-2)
a=5: c=4
Since the second equation is already solved for "y", you can replace "y" by "9" in the other equation. Then solve the new equation for "x".
a2+30a+56=0 , solve for a Using the quadratic formula, you will find that: a=-2 , a=-28
You cannot solve one linear equation in two variables. You need two equations that are independent.
3(5x-2y)=18 5/2x-y=-1