Question

I need someone’s Help on this question

Answer 1

Question 1:

In order to eliminate the x variables in this systems of equations problem using the elimination method, the value of x in the top and bottom equations must cancel each other out.

In order to get this to happen, we can multiply the top equation by 3 and the bottom equation by -4. This way the x value of the top equation becomes 12x and the x value of the bottom equation becomes -12x. These values would cancel each other out, letting you solve for y.

So, question 1 answer: (3, -4) OR (-3, 4)

(both will result in getting a 12x and -12x in the top or bottom that will cancel)

Question 2:

The idea from question 1 can be applied here too:

To eliminate the y variables, we must make the y values in the top and bottom equations cancel each other out.

We can multiply the top equation by 3 and the bottom equation by -7, resulting in -21y in the top equation, and 21y in the bottom equation.

So, question 2 answer: (3, -7) OR (-3, 7)

(both will result in getting a 21y and -21y in the top or bottom that will cancel)

Answer 2

`\begin{array}{llll} \text{\LARGE -3}(4x-7y&=&2)\\ ~~ \text{\LARGE 4}(3x-3y&=&6) \end{array}\implies \stackrel{ \textit{\LARGE eliminating`

now to get them hmmm we can use either equation, say hmmm let's use the 2nd one

`9x+0=36\implies x=\cfrac{36}{9}\implies \boxed{x=4} \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{4(4)-7y=2}\implies 16-7y=2\implies 16=7y+2 \\\\\\ 14=7y\implies \cfrac{14}{7}=y\implies \boxed{2=y} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill~(4~~,~~2)~\hfill~`

quick clarification for a misleading material in the question

what's the second pair? well, is a system of equations of two lines, they only meet at one point, so there's only one pair, or we can say the second pair is the same as the first pair, redundant and unnecessary, but hmmm so it's.