SOLUTION: (3,_),(_,-1),2x-3y=5

Question 70587This question is from textbook Intermediate Algerba
: (3,_),(_,-1),2x-3y=5 This question is from textbook Intermediate Algerba

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
(3,_),(_,-1),2x-3y=5
.
This is two separate problems. The first problem is that you are given (3,_) and the equation
.
2x - 3y = 5
.
The 3 in (3,___) represents x. Substitute 3 for x in the equation and solve for y:
.
2(3) - 3y = 5
.
Multiply out the left side to get:
.
6 - 3y = 5
.
Subtract 6 from both sides:
.
-3y = -1
.
Divide both sides by -3 to end up with:
.
y = (-1/-3) = 1/3
.
So the missing entry in (3,_) is 1/3 and the answer becomes (3, 1/3)
.
The second problem involves the answer set (_, -1) for the equation 2x - 3y = 5. Substitute
-1 for y in the equation to get:
.
2x - 3(-1) = 5
.
And multiply out to get:
.
2x + 3 = 5
.
Subtract 3 from both sides:
.
2x = 2
.
Divide both sides by 2 and the result is:
.
x = 1
.
So the missing x value in (_, -1) is 1 and the answer set is (1, -1)
.
Hope this helps you to understand what the problem was looking for.

RELATED QUESTIONS
(2x^3y^-1)^3 (answered by checkley77)

-1/2x + -3y =... (answered by ikleyn)

solve by elemination method: 5/2x+5/3y=25... (answered by Fermat)

Solve the system 2(2x+3y)=5... (answered by lynnlo)

(2x^2/3y^5)^-3 (answered by stanbon)

-1/3(3y+5) (answered by tommyt3rd)

solve each system by any method 1 x-3y=1 3x-5y=-5 2 2x-3y=5... (answered by jim_thompson5910)

3x+5=2x+2 4x+3=5x-4 2x-5=4x-1 5x+24=2x+15... (answered by CubeyThePenguin)

What is... (answered by edjones)