Question

Whats the solutions to

Answer 1

X+2y=4
Multiply both side by 2
x*2+2y*2=4*2
2x+4y=8

2x-2y=5
Next, use substitute property
2x+4y=8
-
2x-2y=5
=
6y=3
Divided both side by 6
6y/6=3/6
y=0.5

x+2y=4
Substitute y with 0.5
x+2(0.5)=4
x+1=4
Subtract 1 to each side
x+1-1=4-1
x=3, so the solution to this equation is (3,0.5)

7x-2y=-20
Multiply both side by 2
7x*2-2y*2=-20*2
14x-4y=-40

9x+4y=-6
Then, Use substitute property
14x-4y=-40
+
9x+4y=-6
=
23x=-46
Divided 23 to each side
23x/23=-46/23
x=-2

7x-2y=-20
Substitute x with -2
7(-2)-2y=-20
-14-2y=-20
Add 14 to each side
-14-2y+14=-20+14
-2y=-6
Divided -2 to each side
-2y/-2=-6/-2
y=3, so the solution to this equations are (-2,3)

2x+y=7
-3x+y=2
Then, use substitute property
2x+y=7
-
-3x+y=2
=
5x=5
Divided 5 to each side
5x/5=5/5
x=1

2x+y=7
Substitute x with 1
2(1)+y=7
2+y=7
Subtract 2 to each side
2+y-2=7-2
y=5, so the order pair is (1,5). Hope it help!

Answer 2

1st problem:
Solve for x, then substitute that in for x in the second equation. Solve for y. Substitute value of y into first equation. Solve for x. This gives the point of intersection.
$\left \{ {{x+2y=4} \atop {2x-2y=5}} \right. \\ \\ x=4-2y \\ 2(4-2y)-2y=5 \\ 8-8y-2y=5 \\ -10y=-3 \\ y=\frac{3}{10} \\ x=4-2(\frac{3}{10} )= \frac{23}{5} \\ ( \frac{23}{5}, \frac{3}{10} )$

2nd problem:
Multiply the first equation by 2 and then add the two equations. Solve for x. Substitute the value for x into the second equation. Solve for y. This gives the point of intersection.
$\left \{ {{7x-2y=-20} \atop {9x+4y=-6}} \right. \\ \\ 14x-4y=-40 \\ 23x=-46 \\ x=-2 \\ 7(-2)-2y=-20 \\ -2y=-34 \\ y=17 \\ (-2,17)$

3rd problem:
Solve for y in the first equation and substitute that into the second equation. Solve for x. Substitute into first equation. This gives the point of intersection.
$\left \{ {{2x+y=7} \atop {-3xy=2}} \right. \\ \\ y=7-2x \\ -3x+(7-2x)=2 \\ -5x=-5 \\ x=1 \\ y=7-2(1)=5 \\ (1,5)$