Answer:
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Step-by-step explanation:
 
        
             
        
        
        

Solve the following using Substitution method
2x – 5y = -13
3x + 4y = 15


- To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

- Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

- Add 5y to both sides of the equation.


- Multiply  times 5y - 13. times 5y - 13.

- Substitute  for x in the other equation, 3x + 4y = 15. for x in the other equation, 3x + 4y = 15.

- Multiply 3 times  . .

- Add  to 4y. to 4y.

- Add  to both sides of the equation. to both sides of the equation.

- Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

- Substitute 3 for y in  . Because the resulting equation contains only one variable, you can solve for x directly. . Because the resulting equation contains only one variable, you can solve for x directly.


- Add  to to by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible. by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

- The system is now solved. The value of x & y will be 1 & 3 respectively.

 
        
             
        
        
        
When x = -2, y = -5
When x = -1, y = -4
When x = 0, y = -3
When x = 1, y = -2
When x = 2, y = -1
Now plot these points on the graph: (-2,-5), (-1,-4), (0,-3), (1,-2), (2,-1)
        
             
        
        
        
1. -6 ≤ x < -1 . . . . conjunction
2. x ≤ 6 . . or . . 10 ≤ x . . . . disjunction
3. 7 ≤ x < 12 . . . . conjunction
4. x < -9 . . or . . -3 ≤ x . . . . disjunction
5. 2 ≤ x ≤ 5 . . . . conjunction
6. x ≤ 54 . . or . . 66 ≤ x . . . . disjunction
7. 39 < x ≤ 43 . . . . conjunction
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Your problem statement provided no letters.