9x-6y-(12*2y)=0 9x-6y-24y=0 9x-30y=o 9x=30y 3x=10y y=0.3x x=10/3y
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.
Multiplication problems can be solved by consulting a multiplication table. Large numbers can be multiplied using a technique called Long Multiplication. One can also use an electronic calculator.
Multiply every term in both equations by any number that is not 0 or 1, and has not been posted in our discussion already. Then solve the new system you have created using elimination or substitution method:6x + 9y = -310x - 6y = 58
14x6=84+2=86
5x - 4y ≥ -203x - 2y ≤ -8y ≥ -3
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.
Multiplication problems can be solved by consulting a multiplication table. Large numbers can be multiplied using a technique called Long Multiplication. One can also use an electronic calculator.
0.09p=0117 0.09p=0117
Multiply every term in both equations by any number that is not 0 or 1, and has not been posted in our discussion already. Then solve the new system you have created using elimination or substitution method:6x + 9y = -310x - 6y = 58
Elimination is particularly easy when one of the coefficients is one, or the equation can be divided by a number to reduce a coefficient to one. This makes substitution and elimination more trivial.
X=0 and y=0
14x6=84+2=86
You cannot solve one linear equation in two variables. You need two equations that are independent.
5x - 4y ≥ -203x - 2y ≤ -8y ≥ -3
x+y=2 x-4=4
To solve systems of equations using elimination, first align the equations and manipulate them to eliminate one variable. This is often done by multiplying one or both equations by suitable constants so that the coefficients of one variable are opposites. After adding or subtracting the equations, solve for the remaining variable, then substitute back to find the other variable. For inequalities, the same elimination process applies, but focus on determining the range of values that satisfy the inequalities.
I need to solve this problem using the multiplication principle. -5/6n=-10 solution: -5/6n=-10 (- gets canceled) 5/6n=10 ½=6n N=1/6*2 N=1/12