A system of equations is a collection of two or more equations with the same variables. When solving this system, u need to find the unknown variables. One way of solving a system of equations is by substitution.
example :
2x + 2y = 6 (equation 1)
3x + y = 4 (equation 2)
we need to pick a variable and isolate it. The easiest one to pick since it is already by itself is the y in the second equation.
3x + y = 4.....subtract 3x from both sides
y = -3x + 4
now we can sub -3x + 4 in for y in the 1st equation...u have to make sure u sub it back into the 1st equation and not the same equation u used to find it.
2x + 2y = 6.....sub in -3x + 4 in for y and solve for x
2x + 2(-3x + 4) = 6...distribute thru the parenthesis
2x - 6x + 8 = 6...subtract 12 from both sides
2x - 6x = 6 - 8...simplify
-4x = -2...divide both sides by -4
x = -2/-4
x = 1/2
now that we have a numerical number for x, u can sub this back into either of ur equations to find a numerical answer for y.
y = -3x + 4...when x = 1/2
y = -3(1/2) + 4
y = -3/2 + 4
y = -3/2 + 8/2
y = 5/2
so ur solution is : (1/2,5/2) <===
and u can check ur answers by subbing them into ur equations to see if they satisfy both equations...because for it to be a solution to this system, it has to satisfy both equations and not just one of them
Answer:
The Quadratic Polynomial is
2 x² +x -4=0
Using the Determinant method to find the roots of this equation
For, the Quadratic equation , ax²+ b x+c=0
(b) x²+x=0
x × (x+1)=0
x=0 ∧ x+1=0
x=0 ∧ x= -1
You can look the problem in other way
the two Quadratic polynomials are
2 x²+x-4=0, ∧ x²+x=0
x²= -x
So, 2 x²+x-4=0,
→ -2 x+x-4=0
→ -x -4=0
→x= -4
∨
x² +x² +x-4=0
x²+0-4=0→→x²+x=0
→x²=4
x=√4
x=2 ∧ x=-2
As, you will put these values into the equation, you will find that these values does not satisfy both the equations.
So, there is no solution.
You can solve these two equation graphically also.
Ralph hires 8 workers
What is variable cost ?
Variable costs are expenses that alter as the volume of a good or service a company produces fluctuates. Marginal costs multiplied by the number of units produced make up variable costs. They can be regarded as typical expenses as well. Total cost is divided into two parts: fixed costs and variable costs.
Workers paid = $100
Assuming $1600 is given as a Total cost for the 2000 quantity for chicken and Fixed cost is $800
Total cost = $1600 out of which $800 is the fixed cost
Variable cost = Total cost - Fixed cost
= 1600 - 800
= 800
Numbers of workers hired = Variable cost / Workers pay
Numbers of workers hired = 800 / 100
Numbers of workers hired = 8
Therefore, the number of workers that Ralph hires is 8 workers
To learn more about variable cost from the given link
brainly.com/question/14313017
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Answer:
Last one
Step-by-step explanation:
Use the distributive property to multiply 1/5 by 4-3x.
Multiply 1/5 and 4 to get 4/5.
Multiply 1/5 and -3 to get -3/5.
Use the distributive property to multiply 1/7 by 3x-4.
Multiply 1/7 and 3 to get 3/7 & 1/7 × -4 to get -4/7.
Subtract 
from both sides.
Combine 
and 
to get 
.
Subtract 4/5 from both sides.
The least common multiple of 7 and 5 is 35. Convert -4/7 and 4/5 to fractions with denominator 35.
Because 
and 
have the same denominator, subtract them by subtracting their numerators.
Subtract 28 from -20 to get -48.
Multiply both sides by 
, the reciprocal of 
.
Multiply 
by by multiplying the numerator by the numerator and the denominator by the denominator.
Carry out the multiplications in the fraction 
.
Reduce the fraction 1680/1260 to its lowest terms by extracting and cancelling out 420.
