(a) Using substitution, we can solve for one variable in terms of the other and substitute into the other equation. For example, from the first equation, we have x = 4y - 15. Substituting this into the second equation, we get -2(4y-15) + y = -19, which simplifies to -8y + 30 + y = -19, or -7y = -49, or y = 7. Substituting this back into the first equation, we get x = 4(7) - 15 = 13. Therefore, the solution is (13, 7).
(b) Using elimination, we can multiply the second equation by 2 and add it to the first equation to eliminate x. We get 4x - 10y + (-4x + 10y) = 42 + 2(17), simplifying to 0 = 76, which is false. Therefore, there is no solution.
(c) Using elimination, we can multiply the first equation by -4 and add it to the second equation to eliminate x. We get 20x + 12y + (-20x + 36y) = 36 - 108, simplifying to 0 = -72. This is also false. Therefore, there is no solution.
(d) Using substitution, we can solve for x in terms of y from the second equation. We get x = (2/3)y + (2/3). Substituting this into the first equation, we get (1/2)((2/3)y + (2/3)) + 2y = 2, which simplifies to (1/3)y + (1/3) + 2y = 2, or (7/3)y = 5/3, or y = (5/3)(3/7) = 5/7. Substituting this back into the expression for x, we get x = (2/3)(5/7) + (2/3) = 4/7. Therefore, the solution is (4/7, 5/7).
Solve each system of equations using any of the following methods: substitution,
elimination, echelon, or Gauss-Jordan. If there are no solutions, say so. If there are an
infinite number of solutions, parametrize the answer using y as the parameter.
(a) x − 4y = −15
−2x + y = −19
(b) 4x − 10y = 42
−2x + 5y = 17
(c) 15x + 9y = 27
20x + 12y = 36
(d) 1
2
x + 2y = 2
−2x +
3
4
y = 1
1 answer
1 answer