Step 1) Solve the second equation for yy:
2x + y = 32x+y=3
2x + y - color(red)(2x) = 3 - color(red)(2x)2x+y−2x=3−2x
2x - color(red)(2x) + y = 3 - 2x2x−2x+y=3−2x
0 + y = 3 - 2x0+y=3−2x
y = 3 - 2xy=3−2x
Step 2) Substitute 3 - 2x3−2x for yy in the first equation and solve for xx:
3x + 2y = 43x+2y=4 becomes:
3x + 2(3 - 2x) = 43x+2(3−2x)=4
3x + (2 xx 3) - (2 xx 2x) = 43x+(2×3)−(2×2x)=4
3x + 6 - 4x = 43x+6−4x=4
3x - 4x + 6 = 43x−4x+6=4
(3 - 4)x + 6 = 4(3−4)x+6=4
-1x + 6 = 4−1x+6=4
-x + 6 - color(red)(6) = 4 - color(red)(6)−x+6−6=4−6
-x + 0 = -2−x+0=−2
-x = -2−x=−2
color(red)(-1) xx -x = color(red)(-1) xx -2−1×−x=−1×−2
x = 2x=2
Step 3) Substitute 22 for xx in the solution to the second equation at the end of Step 1 and calculate yy:
y = 3 - 2xy=3−2x becomes:
y = 3 - (2 xx 2)y=3−(2×2)
y = 3 - 4y=3−4
y = -1y=−1
The solution is x = 2x=2 and y = -1y=−1 or (2, -1)(2,−1)
