In this problem, solve the system of equations using substitution or elimination.?
1. x - 3y +5 = 0
2x +3y - 5= 0
1. x - 3y +5 = 0
2x +3y - 5= 0
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Given:
x - 3y + 5 = 0
2x + 3y - 5 = 0
Manipulate the equations
x - 3y + 5 = 0; subtract 5 from both sides
x - 3y + 5 - 5 = 0 - 5; solve
x - 3y = -5
2x + 3y - 5 = 0; add 5 to both sides
2x + 3y - 5 + 5 = 0 + 5; solve
2x + 3y = 5
Put the two new equations together
x - 3y = -5
2x + 3y = 5 (+)
__________
3x = 0; divide both terms by 3
3x/3 = 0/3; solve
x = 0
Use substitution to solve for y
0 - 3y + 5 = 0; combine like terms
-3y + 5 = 0; subtract 5 from both sides
-3y + 5 - 5 = 0 - 5; simplify
-3y = -5; divide both sides by -3
-3y/-3 = -5/-3; solve
y = 5/3 <-----------Answer
Check:
0 - 3(5/3) + 5 = 0?
0 - 15/3 + 5 = 0?
0 - 5 + 5 = 0?
0 = 0; it checks
2(0) + 3(5/3) - 5 = 0?
0 + 15/3 - 5 = 0?
0 + 5 - 5 = 0?
5 - 5 = 0?
0 = 0; it checks
Blessings
x - 3y + 5 = 0
2x + 3y - 5 = 0
Manipulate the equations
x - 3y + 5 = 0; subtract 5 from both sides
x - 3y + 5 - 5 = 0 - 5; solve
x - 3y = -5
2x + 3y - 5 = 0; add 5 to both sides
2x + 3y - 5 + 5 = 0 + 5; solve
2x + 3y = 5
Put the two new equations together
x - 3y = -5
2x + 3y = 5 (+)
__________
3x = 0; divide both terms by 3
3x/3 = 0/3; solve
x = 0
Use substitution to solve for y
0 - 3y + 5 = 0; combine like terms
-3y + 5 = 0; subtract 5 from both sides
-3y + 5 - 5 = 0 - 5; simplify
-3y = -5; divide both sides by -3
-3y/-3 = -5/-3; solve
y = 5/3 <-----------Answer
Check:
0 - 3(5/3) + 5 = 0?
0 - 15/3 + 5 = 0?
0 - 5 + 5 = 0?
0 = 0; it checks
2(0) + 3(5/3) - 5 = 0?
0 + 15/3 - 5 = 0?
0 + 5 - 5 = 0?
5 - 5 = 0?
0 = 0; it checks
Blessings
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substitution:
x-3y+5=0---------(1)
2x+3y-5=0------------(2)
x=3y-5---------------(3)
substitute (3)into (1) /cannot substitute (3)into (2) cause it's because (3) is derived from (2)
2(3y-5)+3y-5=0
6y-10+3y-5=0
9y=15
y=15/9
sub y=15/9 into (1)
x-3(15/9)+5=0
x-5+5=0
x=0
i think you have typed wrongly! do you mean 2. x+3y-5??
for elimination for 2.x+3y-5
(x-3y+5=0)x2
- 2x-6y+10=0
2x+3y-5=0
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try to do yourself
x-3y+5=0---------(1)
2x+3y-5=0------------(2)
x=3y-5---------------(3)
substitute (3)into (1) /cannot substitute (3)into (2) cause it's because (3) is derived from (2)
2(3y-5)+3y-5=0
6y-10+3y-5=0
9y=15
y=15/9
sub y=15/9 into (1)
x-3(15/9)+5=0
x-5+5=0
x=0
i think you have typed wrongly! do you mean 2. x+3y-5??
for elimination for 2.x+3y-5
(x-3y+5=0)x2
- 2x-6y+10=0
2x+3y-5=0
-----------------
----------------
try to do yourself
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y= 5/3 and x= 0 or (0,5/3)
To use elimination, the top equation should be multiplied by -2. After adding them together, that leaves you with y= 5/3. If you put it into any of the two given equations you get zero for x.
To use elimination, the top equation should be multiplied by -2. After adding them together, that leaves you with y= 5/3. If you put it into any of the two given equations you get zero for x.
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you know, this is cheating, and you just don't learn. you'll find yourself at a car wash at the age of 25 w/ nothing else to do, and by 40, you'll be irreversibly fcked up
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because the equations are ready for elimination, i would go for it. substitution is harder but it would give out a result
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i presume at the start of the question it says x=example 1 and y=example 2
so you would have plus 1-3(2)=-5 plus 5 = 0
hope i helped
so you would have plus 1-3(2)=-5 plus 5 = 0
hope i helped
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-x + 3y = 5
2x - 3y = 5------ADD
x = 10
- 10 + 3y = 5
3y = 15
y = 5
x = 10 , y = 5
2x - 3y = 5------ADD
x = 10
- 10 + 3y = 5
3y = 15
y = 5
x = 10 , y = 5