Math problems?

Question

Solve the system using addition method:
3x-y=-5
2x+5y=-9

Solve the system using subtraction method:

y=7x-5
2x+3y=8

What is the process to solve these?

oh sorry I meant substitution for the 2nd one

Answer 1

Addtion method
3x-y = -5 (Eq.1)
2x + 5y = - 9 (eq. 2)

Multiply Eq. 1 by 5
5(3x-y = -5)
15x - 5y = -25 (Eq. 3)
Add Eq. 3 and Eq. 2
(15x - 5y) + (2x +5y) = -25 + (-9)
17x = -34
17x/17 = -34/17
x = -2 ===> value of x
Substitute the value of x to Eq. 2
2(-2) + 5y = - 9
-4 + 5y = - 9
5y = - 9 + 4
5y = -5
5y/5 = -5/5
y = -1 ===> value of y
To check, subtitute the value of x and y to Eq. 1
3(-2) - (-1) = - 5
-6 + 1 = - 5
-5 = - 5
Subtraction method
y = 7x - 5 (Eq. 1)
2x + 3y = 8 (Eq. 2)
Multiply Eq. 1 by 3
3(y = 7x - 5)
3y = 21x - 15 (Eq. 3)
Subtract Eq. 2 from Eq. 3

3y = 21x - 15 ( Eq. 3)
-(2x + 3y = 8) (Eq. 2) =

3y = 21x - 15
-2x -3y = -8
____________
-2x = 21x - 23

-2x - 21x = 21x - 21x - 23
-23x = -23
-23x/-23 = -23/-23
x = 1 ==> value of x
Substitute the value of x to Eq. 1
y = 7(1) - 5
y = 7 - 5
y = 2 ==> value of y
To check, substitute the value of x and y to Eq. 2
2(1) + 3(2) = 8
2 + 6 = 8
8 = 8

Answer 2

the PROCESSES for these:
"the addition method"
you want either the "x" terms or the "y" terms to cancel out. the way to do this is to multiply the equations by certain numbers so that in one equation the variable ("x" or "y") is equal to that in the second equation but one of them is negative.
(that was kinda hard to understand so heres how to do it with ur example:)

3x-y=5
2x+5y=9

we will multiply each term in the first equation by 5.
now we have:

15x - 5y = 25
2x +5y = 9

if you add these up you'll see that the X's make "17x" (bc 15x+2x = 17x) and the y's make zero. (bc 5y - 5y = 0) and the number equals 34 (bc 25+9=34)

so you have one equation that is:

17x = 34
now you divide by 17 on both sides in order to solve for x.
now you find that x = 2.

once u know that x = 2 you go back to one of the origional equations (it doesnt matter which one) and plug in "2" for "x" and then solve for y.
if we do that with the first equation we get:
3(2)-y=5
so 6-y=5
so -y=-1
and y=1

done.


now the substitution method:

y=7x-5
2x+3y=8

since we know that "y" = "7x-5" (bc its given to us in the equation) so we just plug in "7x-5" for "y" in the second equation and get:

2x + 3(7x-5) = 8
now distribute the 3 into the parenthesis:

2x + 21x - 15 = 8
if you solve for x now you'll get:

23x = 23
x = 1.

now we can solve for "y" by plugging in "1" for "x" in any of the origional equations:

if we take the first one:
y=7x-5
so y=7-5
so y=2

(usually you would use the substitution method if one of the equations is set up "y=...." or "x=...") (but if the x's and y's are on the same side and each have numbers in front of them you would usually use the addition method.)

done.

good luck.

Answer 3

15x - 5y = 25 (after X by 5)
x + 5y = - 9------ADD
16x = 16
x = 1
1 + 5y = - 9
5y = - 10
y = - 2
Answer x = 1,y = - 2
(obtained by taking second equation as
x + 5y = - 9 as opposed to the 2x + 5y = - 9 as given)


2x + 3(7x - 5) = 8
2x + 21x - 15 = 8
23x = 23
x = 1
y = 7 - 5 = 2
Answer is x = 1, y = 2
(obtained by SUBSTITUTION method.In this case substituted y = 7x - 5 into 2nd equation)

Answer 4

addition and subtraction?

top one x = -2, y = -1
bottom one x=1, y = 2