Question

determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.
$-x + 6y = 7$
$6x - 30y = -34$
answer attempt 1 out of 2
$circ$ infinitely many solutions
$circ$ no solutions
$circ$ one solution
submit answer

Explanation:

Step1: Isolate x from first equation

From $-x + 6y = 7$, rearrange to get $x = 6y - 7$.

Step2: Substitute x into second equation

Substitute $x = 6y - 7$ into $6x - 30y = -34$:
$$6(6y - 7) - 30y = -34$$
Expand and simplify:
$$36y - 42 - 30y = -34$$
$$6y - 42 = -34$$
$$6y = 8$$
$$y = \frac{8}{6} = \frac{4}{3}$$

Step3: Solve for x using y value

Substitute $y = \frac{4}{3}$ into $x = 6y - 7$:
$$x = 6\times\frac{4}{3} - 7 = 8 - 7 = 1$$

Step4: Verify solution consistency

Check if $x=1, y=\frac{4}{3}$ satisfies both equations:

1. $-1 + 6\times\frac{4}{3} = -1 + 8 = 7$ (matches first equation)
2. $6\times1 - 30\times\frac{4}{3} = 6 - 40 = -34$ (matches second equation)

A unique consistent solution exists.

Answer:

One Solution