Can someone assit me with some Algebra problems?

Question

1.Bob invested $21,000, part at 18% and part at 8%. If the total interest at the end of the year is $1,980, how much did he invest at 18%?

2.Solve the system by addition.
–3x + 4y = –16
x – 2y = 12
3.Graph f(x) = 3x + 2.

4.Solve the following system of linear inequalities by graphing.
x – 2y less than equal to 4
x greater than equal to 4

Answer 1

1. .18x+.08(21,000-x)=1980 Now, you turn the crank.
2. x-y=4
3. Slope=3, y-intercept=2
4. I don't do pictures

Answer 2

1) x + y=21000
y= 21000-x

1980= (18%x X) + 8%x (21000-x)
1980= 0.18x + 1680 - 0.08x
300= 0.18x- 0.08x
30 000= 18x-8x
10x= 30 000
x= 3000

therefore Bob invested $3000 at 18%

2) -3x+4y=-16
x-2y=12

x=12+2y

-3x+4y=-16
-3(12+2y)+4y= -16
-36- 6y+ 4y= -16
-2y= 20
y= -10

therefore x= 2y +12
x= -20+12
x= -8

3) f(x)=3x+2

Intercepts Y axis at 2
Intercepts X axis at 2/3


4) x- 2y less than equal to 4
x greater thann equal to 4

Draw the line x= 4

Draw the line y= x/2 - 2

The area above the line y= x/2 -2 and past X=4 is the solution

Answer 3

You can AIM "Studybuddy".

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