In the xy=plane, the lines given are 2x-y=3 and x+2y=4, intersect at which points?

Answer 1

2x-y=3
y=2x-3 --- (1)

x+2y=4
x=4-2y --- (2)

sub (1) into (2)

x=4-2(2x-3)
x=4-4x+6
x=10-4x
5x=10
x=2

therefore, when x=2

y=2(2)-3
y=4-3
y=1

lines intersect at (2,1)

Answer 2

Find the intersection of the given lines.

2x - y = 3
x + 2y = 4

Add twice the first equation to the second.

5x = 10
x = 2

Substitute into the second equation.

2 + 2y = 4
2y = 2
y = 1

The point of intersection of the two lines is (2,1).

Answer 3

first equation ==> y = 2x - 3, x = (3 + y)/2
second equation ==> y = (4 - x)/2, x = 4 - 2y
the point where y intersect is
2x - 3 = (4 - x)/2
4x - 6 = 4 - x
5x = 10 ==> x = 2

the point where x intersect is
(3 + y)/2 = 4 - 2y
3 + y = 8 - 4y
5y = 5
y = 1
They intersect at point (2,1)

Answer 4

x = 2, y = 1; (2,1) are the points of intersection. When you do the simultaneous equation solution by multiplying 2 in the first equation and adding the second column, you will get x = 2 and substituting the value of "x" in the second equation and solving for "y" will yield y = 1.

Answer 5

write in standard form

y = 2x - 3
y = -1/2x + 2

set equal to each other and solve for x

2x - 3 = -1/2x + 2

4x - 6 = -x + 4
5x = 10
x = 2

sub into either equation

4 -y = 3

y = 1

the two lines cross at (2,1)

Answer 6

of path, via fact the two equations behave in accordance to their respective slope as defined via their coordinates, Assign X=0 and resolve for Y and vice versa, try this to each linear equation then plot the strains as defined via ( 0,Y) and ( X,0) coordinates. in fact we've discovered from airplane geometry that parallel strains under no circumstances intersect to a minimum of one yet another. as a result all strains in one easy intersect one yet another different than parallel strains.