Solve the linear equations by using substitution -2X-5Y=3 3X+8Y=-6

[tex]\sf -2x-5y=3[/tex]
[tex]\sf 3x+8y=-6[/tex]

Let's solve the second equation for 'x':

[tex]\sf 3x+8y=-6[/tex]

Subtract 8y to both sides:

[tex]\sf 3x=-8y-6[/tex]

Divide 3 to both sides:

[tex]\sf x=-\dfrac{8}{3}y-2[/tex]

Now let's plug this in for 'x' in the first equation:

[tex]\sf -2x-5y=3[/tex]

[tex]\sf -2(-\dfrac{8}{3}y-2)-5y=3[/tex]

Distribute:

[tex]\sf \dfrac{16}{3}y+4-5y=3[/tex]

Combine like terms:

[tex]\sf \dfrac{1}{3}y+4=3[/tex]

Subtract 4 to both sides:

[tex]\sf \dfrac{1}{3}y=-1[/tex]

Divide 1/3 to both sides or multiply by its reciprocal, 3:

[tex]\sf y=-3[/tex]

This is the y-value of our solution, we can plug it into any of the two equations to find the x-value:

[tex]\sf 3x+8y=-6[/tex]

[tex]\sf 3x+8(-3)=-6[/tex]

Multiply:

[tex]\sf 3x-24=-6[/tex]

Add 24 to both sides:

[tex]\sf 3x=18[/tex]

Divide 3 to both sides:

[tex]\sf x=6[/tex]

So our final solution is:

[tex]\boxed{\sf (6,-3)}[/tex]