Answer to Question #305273 in Linear Algebra for Nhlonipho Manganyi

Question

Consider the linear equation 2a + 3b = 4

Is (a; b) = (1; 1

2

) a solution to the equation? Motivate your answer.

Answer 1

If $(a, b)=(1,\dfrac{1}{2}),$ then substitute

$$2(1) + 3(\dfrac{1}{2}) = 4$$

$$\dfrac{5}{2}=4, False$$

Therefore $(a, b)=(1, \dfrac{1}{2})$ is not a solution to the equation $2a+3b=1.$

If $(a, b)=(\dfrac{1}{2}, 1),$ then substitute

$$2(\dfrac{1}{2}) + 3(1) = 4$$

$$4=4, True$$

Therefore $(a, b)=(\dfrac{1}{2}, 1)$ is a solution to the equation $2a+3b=1.$