The easiest way to solve this question is to use substitution. Since $\displaystyle y=7x-2$ you can replace y for 7x-2 in the other equation.

You should have

$\displaystyle 2(7x-2)=6x+4$.

Distribute the 2 to the parentheses.

$\displaystyle 14x-4=6x+4$

Add 4 to both sides of the equation.

$\displaystyle 14x=6x+8$

Subtract 6x from both sides.

$\displaystyle 8x=8$

Divide by 8 to get x.

$\displaystyle x=1$

Put 1 back in to either equation for x to solve for y.

$\displaystyle y=3(1)+2$

$\displaystyle y=5$

First task is to solve at least one of the equations for y.

$\displaystyle -3x+2y=19$

Move -3x to the other side by adding 3x to both sides.

$\displaystyle 2y=3x+19$

Divide by 2 to all the terms in the equation.

$\displaystyle y=\frac{3}{2}x+\frac{19}{2}$

Plug this value for y into the other equation.

$\displaystyle -3x-19=2(\frac{3}{2}x+\frac{19}{2})$

Distribute the 2.

$\displaystyle -3x-19=3x+19$

Add 3x to both sides.

$\displaystyle -19=6x+19$

Subtract 19 from both sides of the equation.

$\displaystyle -38=6x$

Divide by 6.

$\displaystyle x=-6.33 or -6\frac{1}{3}$

Plug this back in for x in either equation.

$\displaystyle -3(-6.33)+2y=19$

$\displaystyle 19+2y=19$

$\displaystyle 2y=0$

$\displaystyle y=0$

To solve this system of equations, you are given the value of $\displaystyle y$.

$\displaystyle y = 2x - 16$

The second equation is $\displaystyle y + 3x = 9$

So you put the value of $\displaystyle y$ into the second equation.

$\displaystyle 2x-16 + 3x = 9$

Combine like terms.

$\displaystyle 5x -16 = 9$

Add $\displaystyle 16$ to both sides of the equation.

$\displaystyle 5x - 16 + 16 = 9 + 16$

$\displaystyle 5x = 25$

Divide both sides by $\displaystyle 5$.

$\displaystyle \frac{5x}{5} = \frac{25}{5}$

$\displaystyle x = 5$

Substitute the value of x in one of the equations to get the value of y.

$\displaystyle y = 2(5) - 16$

$\displaystyle y = 10-16$

$\displaystyle y = -6$

$\displaystyle (5,-6)$ is the correct answer.

$\displaystyle x + y = 5$

$\displaystyle 2x-3y = 5$

To cancel out the $\displaystyle x$ terms, multiply $\displaystyle x + y = 5$ by $\displaystyle -2$

$\displaystyle -2x - 2y = -10$

Then add:

 $\displaystyle +$$\displaystyle 2x-3y = 5$

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           $\displaystyle -5y = -5$

$\displaystyle y = 1$

Plug the value of $\displaystyle y,$ which is $\displaystyle 1$ into one of the equations to get the value of $\displaystyle x$.

$\displaystyle x + 1 = 5$

$\displaystyle x+1-1 = 5 - 1$

$\displaystyle x = 4$

$\displaystyle (4,1)$ is the correct answer.

Using the substitution method, set the two systems of equations equal to each other.

$\displaystyle 6x = x-15$

Isolate the variable by subtracting $\displaystyle x$ from both sides of the equation.

$\displaystyle 6x-x = x-x-15$

$\displaystyle 5x = -15$

$\displaystyle \frac{5x}{5} = \frac{-15}{5}$

$\displaystyle x = -3$

To get the value of $\displaystyle y$, substitute the value of $\displaystyle x$ in one of the equations.

$\displaystyle y = 6(-3)$

$\displaystyle y = -18$

$\displaystyle (-3,-18)$ is the correct answer.

Using the substitution method, set both systems of equations equal to each other.

$\displaystyle -x + 6 = 2x - 3$

Isolate the variable by adding $\displaystyle x$ to both sides of the equation.

$\displaystyle -x + x + 6 = 2x + x -3$

$\displaystyle 6 = 3x - 3$

Add $\displaystyle 3$ to both sides.

$\displaystyle 9 = 3x$

Divide both sides by 3.

$\displaystyle \frac{9}{3} =\frac{3x}{3}$

$\displaystyle 3 = x$

To get the value of y, substitute the value of x in one of the equations.

$\displaystyle 2 (3) - 3 = 6 - 3 = 3$

$\displaystyle y = 3$

$\displaystyle (3,3)$ is the correct answer.

In order to eliminate the $\displaystyle x$ terms, first multiply the first equation by $\displaystyle -4.$

$\displaystyle -4 (-5x+10y) = (-4) 60$

$\displaystyle 20x - 40y = -240$

Then multiply the second equation by $\displaystyle 5.$

$\displaystyle 5 (-4x + 16y) = 5 (32)$

$\displaystyle -20 x +80y = 160$

This will now eliminate the $\displaystyle x$ terms when added together.

       $\displaystyle 20x - 40y = -240$

+ $\displaystyle -20 x +80y = 160$

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                 $\displaystyle 40y = -80$

Divide both sides by $\displaystyle 40.$

$\displaystyle y = -2$

Now substitute the value of $\displaystyle y,$ which is $\displaystyle -2,$ to get the value of  in one of the the two original equations.

$\displaystyle -5x-20 = 60$

Add $\displaystyle 20$ to both sides of the equation.

$\displaystyle -5x -20 + 20 = 60 + 20$

$\displaystyle -5x = 80$

Divide both sides by $\displaystyle -5.$

$\displaystyle 80 \div -5 = -16$

$\displaystyle x = -16$

$\displaystyle (-16,-2)$ is the correct answer.

Set both equations equal to each other and solve.

$\displaystyle -5x + 2 = 3x - 14$

Add $\displaystyle 5x$ to both sides of the equation.

$\displaystyle -5x + 5x + 2 = 3x + 5x - 14$

$\displaystyle 2 = 8x - 14$

Add $\displaystyle 14$ to both sides of the equation.

$\displaystyle 2+14 = 8x-14+14$

$\displaystyle 16 = 8x$

Divide both sides by 8.

$\displaystyle x = 2$

Plug the value of the $\displaystyle x$ variable into one of the equations to get the value of $\displaystyle y.$

$\displaystyle -5(2) + 2 = y$

$\displaystyle -10 +2 =-8$

$\displaystyle y = -8$

$\displaystyle (2,-8)$ is the correct answer for this system of equations.

To solve this system of equations, multiply the first equation by $\displaystyle 3$.

$\displaystyle 3 (3.5x + 2.5y) =38 (3)$

$\displaystyle 10.5x +7.5y = 114$

Now add the two equations together to remove the $\displaystyle y$ terms.

   $\displaystyle 10.5x +7.5y = 114$

+   $\displaystyle 1.5x -7.5y = -18$

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  $\displaystyle 12.0x$                $\displaystyle = 96$

Divide both sides by $\displaystyle 12$.

$\displaystyle 96\div12 = 8$

$\displaystyle x = 8$

Plug the value of the $\displaystyle x$ variable, which is $\displaystyle 8$ into one of the equations.

$\displaystyle 3.5(8) + 2.5y = 38$

$\displaystyle 28 + 2.5y = 38$

Subtract $\displaystyle 28$ from both sides of the equation.

$\displaystyle 28-28 + 2.5y = 38-28$

$\displaystyle 2.5y = 10$

Divide both sides by $\displaystyle 2.5$

$\displaystyle y = 4$

$\displaystyle (8,4)$ is the correct answer for this system of equations.

To solve this system of equations:

$\displaystyle -\frac{1}{3}x +\frac{1}{6}y = 2$

Multiply the first equation by $\displaystyle 4$ and the second equation by $\displaystyle -6$.

$\displaystyle 4 (-\frac{1}{2}x -\frac{1}{4}y) = 3(4)$

$\displaystyle -2x -y =12$

$\displaystyle -6 (-\frac{1}{3}x+\frac{1}{6} y) = (-6) 2$

$\displaystyle 2x-y = -12$

Add these two equations; this will remove the $\displaystyle x$ terms.

$\displaystyle -2x -y =12$

+$\displaystyle 2x-y = -12$

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          $\displaystyle -2y = 0$

                $\displaystyle y = 0$

To get the value of $\displaystyle x$, plug the value of $\displaystyle y$, which is $\displaystyle 0$ into one of the equations.

$\displaystyle 2x -y = -12$

$\displaystyle 2x - 0 = -12$

$\displaystyle 2x = -12$

Divide both sides by $\displaystyle 2$

$\displaystyle \frac{2x}{2} = \frac{-12}{2}$

$\displaystyle x = -6$

$\displaystyle (-6,0)$ is the correct answer for this system of equations.