Question

Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.

Answer 1

The given system of equations has exactly one solution (1, ⁴/₃).

Explanation:

Given system of equations:

`\begin{cases}-x+6y=7\\6x-30y=-34\end{cases}`

To solve the given system of equations, multiply all terms of the first equation by 5:

`\implies -x \cdot 5+6y \cdot 5=7 \cdot 5`

`\implies -5x+30y=35`

Add this to the second equation to eliminate the term in y:

`\begin{array}{crcccl}&6x & - & 30y & = &-34\\\vphantom{\frac12}+ & (-5x & + & 30y & = & \;\;\:35)\\\cline{2-6}\vphantom{\frac12}&x&&&=&\;\;\;\;\:1\\\cline{2-6}\end{array}`

Therefore, x = 1.

Substitute the found value of x into the first equation and solve for y:

`\implies -(1)+6y=7`

`\implies 6y=8`

`\implies y=(8)/(6)`

`\implies y=(4)/(3)`

Therefore, the given system of equations has exactly one solution (1, ⁴/₃).

Answer 2

• C) One solution

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Given system:

• - x + 6y = 7
• 6x - 30y = - 34

We can solve it by elimination.

Multiply the first equation by 6 and add up to get:

• -6x + 36y + 6x - 30y = 42 - 34
• 6y = 8
• y = 8/6
• y = 4/3

Without solving for x we can see there is one solution.

The last answer choice is the correct one.