Question 184483
The equations are 1st degree. That means there 
are no powers higher than 1. Two 1st degree
equations can either:
(a) Be 2 parallel straight lines
(b) Be 2 straight lines that intersect at 1 point
(c) Be the same equation written twice
If there're parallel, they're lines with the 
same slope, and have no solution. If they're
not parallel, they must have a solution
I'll put them in the form {{{y = mx + b}}}, where
{{{m}}}= slope. If the slope is the same, there's
no solution
(1) {{{ 2x + 3y = 8}}}
(2) {{{ 3x + 2y = 7}}}
-----------------------
(1) {{{ 2x + 3y = 8}}}
(1) {{{3y = -2x + 8}}}
(1) {{{y = -(2/3)x + (8/3)}}}
-----------------------
(2) {{{ 3x + 2y = 7}}}
(2) {{{2y = -3x + 7}}}
(2) {{{y = -(3/2)x + (7/2)}}}
-----------------------
The slopes are not the same, so there is a solution
To solve, I'll multiply (1) by {{{2}}} and (2) by {{{3}}}
(1) {{{ 4x + 6y = 16}}}
(2) {{{ 9x + 6y = 21}}}
Now subtract (1) from (2)
(2) {{{ 9x + 6y = 21}}}
(1) {{{ -4x - 6y = -16}}}
(3) {{{5x = 5}}}
(3) {{{x = 1}}}
Put this back into either equation to find {{{y}}}
(1) {{{ 4*1 + 6y = 16}}}
(1) {{{6y = 16 - 4}}}
(1) {{{y = 2}}}
This tells me that the point of intersection is (1,2)
Now I'll plot them
{{{ graph( 500, 500, -5, 5, -5, 5, -(2/3)x + 8/3,-(3/2)x + 7/2 ) }}}
For each line, to find the x-intercept, set {{{y=0}}} and
solve for {{{x}}}. Then set {{{x=0}}} and solve for {{{y}}}
for the y-intercept