Respuesta :
Both write the equation in the form of directional .
2x + 3y = 6
3y = - 2x + 6
[tex]y = - \frac{2}{3} x + 2[/tex]
a = - 2/3 b = 2
...............................................................
-3x + 5y = 10
5y = 3x + 10
[tex]y = \frac{3}{5} x + 2[/tex]
a = 3/5 b= 2
.....................................
b ratio is the same , so the two lines intersect the y-axis at P = ( 0,2 ) . This joint solution
2x + 3y = 6
3y = - 2x + 6
[tex]y = - \frac{2}{3} x + 2[/tex]
a = - 2/3 b = 2
...............................................................
-3x + 5y = 10
5y = 3x + 10
[tex]y = \frac{3}{5} x + 2[/tex]
a = 3/5 b= 2
.....................................
b ratio is the same , so the two lines intersect the y-axis at P = ( 0,2 ) . This joint solution
Answer:
The solution to the system of linear equations is (0,2)
Step-by-step explanation:
1. Let´s name the two equations given as:
[tex]2x+3y=6 (Eq.1)\\-3x+5y=10 (Eq.2)[/tex]
2. Solve for y on Eq.1:
[tex]2x+3y=6\\3y=6-2x\\y=\frac{6-2x}{3} (Eq.3)[/tex]
3. Replace Eq.3 in Eq.2:
[tex]-3x+5(\frac{6-2x}{3})=10\\-3x+\frac{30-10x}{3}=10\\-3x+\frac{30}{3}-\frac{10x}{3}=10\\ -3x+10-\frac{10x}{3}=10\\-3x-\frac{10x}{3}=10-10\\-3x-\frac{10x}{3}=0\\\frac{-9x-10x}{3}=0\\ -9x-10x=0\\-19x=0\\x=0[/tex]
4. Replace the value of x on Eq.3:
[tex]y=\frac{6-2(0)}{3}\\y=\frac{6}{3}\\y=2[/tex]
Therefore the solution to the system of linear equations is (0,2)