I need help with a few of practice problems from saxon algebra1?

Question

1) Use the substitution method to solve this system of equations.
s+2t=1
3s+t= -7
2) Choose a method and solve.
6x-y=9
3x+4y=-9
Solve by graphing.
3)The difference between three times a number and another number is 2. Their sum is 6 Write a system of equations. Then solve for the numbers by graphing the equations.
4)Find each product.
(3x+2y)^2
5)Use the quadratic formula to solve.
3x^2-5x+1=0

If you could give me the answer and the process and maybe explain the answer to me in laymans terms id really appreciate it. Thanks.

Answer 1

1) Solving the first equation for s gives s = 1 - 2t. Substituting this into equation 2 gives 3(1-2t) + t = 3 - 5t = -7, and this gives -5t = -10, so t = 2.

2) y = 6x-9 from equation 1, then substitution gives 3x + 4(6x-9) = 27x - 36 = -9, so 27x = 27, so x = 1. Then equation 1 shows 6-y=9, so y=-3.

3) The description gives 3x - y = 2 and 3x + y = 6.

4) (3x+2y)^2 = (3x+2y)(3x+2y) = 9x^2 + 12xy + 4y^2

5) 5 +/- sqr(25 -4*3*1) / (2*3)

Answer 2

1) s=-2t+1 corner s and replace s into second equation

3(-2t+1)+t=-7

-6t+t=-10

t=2 put value back into original equation

s=-2(2)+1=-3

2)

y=6x-9 corner y and replace y in second equation

3x+4(6x-9)=-9

3x+24x=27

x=1 place value x back into original equation

y=6(1)-9=-3

graphing solve is make each equal to y and make them equal to each other

6x-9= -9-3x/4

and y=6x-9 and y=-9-3x/4

24x-36=-9-3x

27x=27

x=1

y=6(1)-9=-3

y=-9-3(1)/4=-3

coordinates of intersection (1,-3)
whatever the values of x discovered

3) 3a-b=2 and a+b=6

a=6-b corner a and replace a in second equation

3(6-b)-b=2

-3b-b=-16

b=4 place b value back into original equation

a=6-4=2

4) I don't really know what they mean here nothis is =, let's expand

9x^2+12xy+4y^2

5) quadratic formula

5+-sqrt(25-12)/2(3)

=5+-sqrt(13)/6

x=[5+sqrt(13)]/6

and

x=[5-sqrt(13)]/6

Answer 3

1
s+2t=1
3s+t=-7
solution
s+2t=1............(1)
3s+t=-7...........(2)
from....(1)
s=1-2t..............(3)
substitute d value of s in ....(2)
3(1-2t)+t=-7
3-6t+t=-7
3-5t=-7
5t=-7-3
5t=-10
t=-10/5
t=-2
substitute d value of t in ....(3)
s=1-2(-2)
s=1+4
s=5
therefore:
s=5 and t=-2

2
6x-y=9.......(1)
3x+4y=-9.....(2)
using elimination method:
1by2
2by4
12x-2y=18........(3)
12x+16y=-36.......(4)
subtract (3) from(4)
18y=-54
y=54/18
y=3
substitute d value of y in ......(1)
6x-y=9
6x-3=9
6x=9+3
6x=12
x=12/6
x=2
therefore: x=2 and y=3