+0

# Find all solutions to

0
153
3
+108

Find all solutions to

ab+2a+2b=-16

ac+2a+2c=26

bc+2b+2c=-14

Feb 18, 2021

#1
+112966
+2

My algebra was very long-winded and I have not finished it but

I used the first one to get a in terms of b       a(b)

then the last one to get b in terms of c          b(c)

then i substituted the b(c)  into the a to get a(b)  in terms of a(c)

Then I substituted this into the middle one and solved for c

etc

There is probably a better way.

Feb 18, 2021
edited by Melody  Feb 18, 2021
#2
+299
+2

Use Simon's favorite factoring trick; it's going to make it so much easier:

$$(a+2)(b+2)=-12\\ (a+2)(c+2)=30\\ (b+2)(c+2)=-10$$

Multiply all the equations to get:

$$((a+2)(b+2)(c+2))^2=3600\\ (a+2)(b+2)(c+2)= \pm 60$$

In the first case where the right-hand side of the equation above is positive, to solve for a, divide $$(a+2)(b+2)(c+2)=60$$ by $$(b+2)(c+2)=-10$$ to get $$a+2=-6 \rightarrow a=-8$$. Do that the same with b and c to get $$b = 0$$ and $$c=-7$$.

In the second case where the right-hand side of the equation is negative, do the exact same thing mentioned above to get $$a = 4, b = -4, c = 3$$.

Therefore, the solutions to this system of the equation are:

$$\boxed{a = -8, b = 0, c = -7 \text{ or } a = 4, b = -4, c = 3}$$

Feb 20, 2021
#3
+112966
+1

That is really cool Textot.

Thanks very much for showing me.

Feb 20, 2021