Question 43043
I will solve only one of them. You have the other as an exercise. First, by substitution. We have:

{{{5x + 2y = 1}}}
{{{x - 3y = 7 }}}

We need to isolate x or y on one side of one equation. Let's try the second, which is easier since the coefficient of x is 1:
When we pass a number to the other side, it's sign is the inverse, so we pass {{{-3y}}} as {{{+3y}}} to get:

{{{x = 7 + 3y}}}

The first expression is now solved, but with 7 + 3y in the place of x.

{{{5(7 + 3y) + 2y = 1}}}
Distributive property:
{{{35 + 15y + 2y = 1}}}
Adding like terms:
{{{35 + 17y = 1}}}
*[invoke linear_equation "y", 17, 35, 1 ]

Now, we have that {{{y = -2}}}, substitute that into 
{{{x = 7 + 3y}}}:

{{{x = 7 + 3y}}}
{{{x = 7 + 3*-2}}}
{{{x = 7 - 6}}}
{{{x = 1}}}

Now, what would happen if we tried the first one in the beginning? We would pass 2y to the right side and the coefficient 5, getting:

{{{x = 5(1 - 2y)}}}

And solving by substituting that into the second equation like above.

Now, by elimination:

When we have two equations, we can add the terms to make another true equation, for example:

{{{4 + 5 = 9}}}
{{{3 + 9 = 12}}}
{{{(4+3) + (5+9) = (9+12)}}}

Verify that the last equation is true. We have:

{{{5x + 2y = 1}}}
{{{x - 3y = 7 }}}

And so we want to get the same coefficient into at least one variable (with changed signs) so we can cancel it to 0. We can bring the oefficient of x in the second equation to -5 by multiplying all terms by -5 (and then {{{1x * -5 = -5x}}}), so let's do that:

{{{5x + 2y = 1}}}
{{{-5x + 15y = -35}}}

Now, we add all terms, just like we did above:

{{{(5x-5x) + (2y+15y) = (1-35)}}}

Notice that 5x and -5x cancelled themselves out, leaving us a simple linear equation.

{{{17y = -34}}}
{{{y = -34/17}}}
{{{y = -2}}}

Plug in y in any of our equations now. Let's try the first, {{{5x + 2y = 1}}}.

{{{5x + 2(-2) = 1}}}
{{{5x -4 = 1}}}
{{{5x = 5}}}
{{{x = 5/5}}}
{{{x = 1}}}