Systems of Equations

To solve using substitution, solve one equation for one variable and substitute it into the other equation. Solving x - y = 1 for x, we get x = y + 1. Substituting this into 2x + 3y = 11, we get 2(y + 1) + 3y = 11. Solving for y, we get y = 2. Substituting y = 2 back into x = y + 1, we get x = 3. Therefore, the solution is (x, y) = (2, 3).

To solve using elimination, multiply one or both equations by a constant to make the coefficients of one variable the same. Multiplying the first equation by 2 and the second equation by 3, we get:
6x + 4y = 14
6x - 9y = 3
Subtracting the second equation from the first, we get:
13y = 11
Solving for y, we get y = 1. Substituting y = 1 back into either of the original equations, we can solve for x. Using the first equation, we get 3x + 2(1) = 7, which gives x = 2. Therefore, the solution is (x, y) = (2, 1).

To solve using graphing, graph both equations on the same coordinate plane and find the point of intersection. Graphing the equations, we can see that they intersect at the point (2, -1). Therefore, the solution is (x, y) = (2, -1).

To solve this system of equations, we can use the substitution method. Solving the first equation for x, we get x = 5 - 2y. Substituting this into the second equation, we get 3(5 - 2y) - y = 1. Solving for y, we get y = 2. Substituting y = 2 back into x = 5 - 2y, we get x = 1. Therefore, the solution is (x, y) = (1, 2).

To solve this system of equations, we can use the elimination method. Multiplying the first equation by 2 and the second equation by 3, we get:
4x + 6y = 14
12x - 6y = 30
Adding the two equations, we get:
16x = 44
Solving for x, we get x = 2. Substituting x = 2 back into either of the original equations, we can solve for y. Using the first equation, we get 2(2) + 3y = 7, which gives y = 1. Therefore, the solution is (x, y) = (2, 1).

To solve this system of equations, we can substitute the first equation into the second equation. Substituting y = x^2 - 1 into y = 2x - 3, we get x^2 - 1 = 2x - 3. Rearranging the equation, we get x^2 - 2x + 2 = 0. Factoring the quadratic equation, we get (x - 2)(x - 1) = 0. Setting each factor equal to zero, we get x = 2 and x = 1. Substituting x = 2 back into the first equation, we get y = 2^2 - 1 = 3. Therefore, the solution is (x, y) = (2, 1).

To solve this system of equations, we can use the substitution method. Solving the second equation for y, we get y = 2x - 1. Substituting this into the first equation, we get 3x + 2(2x - 1) = 5. Solving for x, we get x = 1. Substituting x = 1 back into the second equation, we get y = 2(1) - 1 = 1. Therefore, the solution is (x, y) = (1, 2).

To solve this system of equations, we can use the elimination method. Multiplying the first equation by 2 and the second equation by 3, we get:
4x + 6y = 14
12x - 6y = 30
Adding the two equations, we get:
16x = 44
Solving for x, we get x = 2. Substituting x = 2 back into either of the original equations, we can solve for y. Using the first equation, we get 2(2) + 3y = 7, which gives y = 1. Therefore, the solution is (x, y) = (2, 1).

To solve this system of equations, we can substitute the first equation into the second equation. Substituting y = x^2 - 1 into y = 2x - 3, we get x^2 - 1 = 2x - 3. Rearranging the equation, we get x^2 - 2x + 2 = 0. Factoring the quadratic equation, we get (x - 2)(x - 1) = 0. Setting each factor equal to zero, we get x = 2 and x = 1. Substituting x = 2 back into the first equation, we get y = 2^2 - 1 = 3. Therefore, the solution is (x, y) = (2, 1).

To solve this system of equations, we can use the substitution method. Solving the second equation for y, we get y = 2x - 1. Substituting this into the first equation, we get 3x + 2(2x - 1) = 5. Solving for x, we get x = 1. Substituting x = 1 back into the second equation, we get y = 2(1) - 1 = 1. Therefore, the solution is (x, y) = (1, 2).

To solve this system of equations, we can use the elimination method. Multiplying the first equation by 2 and the second equation by 3, we get:
4x + 6y = 14
12x - 6y = 30
Adding the two equations, we get:
16x = 44
Solving for x, we get x = 2. Substituting x = 2 back into either of the original equations, we can solve for y. Using the first equation, we get 2(2) + 3y = 7, which gives y = 1. Therefore, the solution is (x, y) = (2, 1).

To solve this system of equations, we can substitute the first equation into the second equation. Substituting y = x^2 - 1 into y = 2x - 3, we get x^2 - 1 = 2x - 3. Rearranging the equation, we get x^2 - 2x + 2 = 0. Factoring the quadratic equation, we get (x - 2)(x - 1) = 0. Setting each factor equal to zero, we get x = 2 and x = 1. Substituting x = 2 back into the first equation, we get y = 2^2 - 1 = 3. Therefore, the solution is (x, y) = (2, 1).

To solve this system of equations, we can use the substitution method. Solving the second equation for y, we get y = 2x - 1. Substituting this into the first equation, we get 3x + 2(2x - 1) = 5. Solving for x, we get x = 1. Substituting x = 1 back into the second equation, we get y = 2(1) - 1 = 1. Therefore, the solution is (x, y) = (1, 2).

To solve this system of equations, we can use the elimination method. Multiplying the first equation by 2 and the second equation by 3, we get:
4x + 6y = 14
12x - 6y = 30
Adding the two equations, we get:
16x = 44
Solving for x, we get x = 2. Substituting x = 2 back into either of the original equations, we can solve for y. Using the first equation, we get 2(2) + 3y = 7, which gives y = 1. Therefore, the solution is (x, y) = (2, 1).