Here is your system of equations:
[tex]\left \{ {{2x+3y=-6} \atop {3x-4y=-12}} \right.[/tex]
When solving a system, you can either substitute or eliminate. I am going to eliminate because you can only substitute when there are answer choices.
When eliminating, you need either x or y to be the same or opposite(sign). I am going to eliminate y, whuch means 3y and -4y has to have the same number. To make them the same, I will multiply the two numbers because multiplying two numbers is an easier way to find a common multiple.
[tex]3 \times 4 = 12[/tex]
That means they both have to be changed so that their value is 12.
[tex]2x+3y=-6 \\ \\ 3 \times ?=12 \\ ?=4[/tex]
You have to multiply the first equation by 4.
[tex]2x+3y=-6 \rightarrow 8x+12y=-24[/tex]
For the second equation, I will have to multiply by 3 because:
[tex]3x-4y=-12 \\ \\-4 \times ?=-12 \\?=3[/tex]
Multiply the second equation by 3:
[tex]3x-4y=-12 \rightarrow 9x-12y=-36[/tex]
Here is your new system of equations:
[tex]\left \{ {{8x+12y=-24} \atop {9x-12y=-36}} \right.[/tex]
Add:
[tex]{8x+12y=-24} \\{9x-12y=-36} \\ \\ 17x=-60[/tex]
y is eliminated because when a number is added/subtracted by its opposite, it's cancelled out.
Although the case is different, remember:
NEGATIVES
POSITIVES
Now you need to divide both sides by 17 to leave x alone.
[tex]\frac{17x}{17} =\frac{-60}{17} \\ \\ x=-\frac{60}{17}[/tex]
Since x has a value, that means y also does. That also means the answer to your question is one solution.
Answer:
the correct answer is A. This system has no solutions. It is not possible for two x minus three y to equal both negative six and nine at the same time.
Step-by-step explanation:
i hope this helps
