George and a total of $592 in simplest interest from two separate accounts. In an account earning 6% interest, George invested $1800 more than three times the amount he invested in an account earning 4%. How much did he invests in each account

The amount invested at 4% is $2,200 and the amount invested at 6% is $8,400

Step-by-step explanation:

we know that

The simple interest formula is equal to

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

Let

x ----> the amount invested at 4%

(3x+1,800) -----> the amount invested at 6%

we have that

The interest earned by the account at 4% plus the interest earned by the amount at 6% must be equal to $592

so

The linear equation that represent this situation is

solve for x

therefore

The amount invested at 4% is $2,200 and the amount invested at 6% is $8,400

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)