How do you solve 8a+2b=11 and 5a-6b=25?

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Answer 1 · 2018-05-23T03:33:40

$a = 2, b = - \frac{5}{2}$ $a=2, b=-5/2$

We have the following system:

$8 a + 2 b = 11$ $8a+2b=11$

$5 a - 6 b = 25$ $5a-6b=25$

We want to eliminate one of the variables so we can solve for the other. So let's multiply the first system by $3$ $3$ . We get

$24 a + 6 b = 33$ $24a+6b=33$

$5 a - 6 b = 25$ $5a-6b=25$

Now we can add both systems to get

$29 a = 58$ $29a=58$

Dividing both sides by $29$ $29$ , we get

$a = 2$ $color(blue)(a=2)$

We've solved for one of the variables, now we can plug into an equation to solve for the other.

I'll plug into the first equation. We get

$8 (2) + 2 b = 11$ $8(2)+2b=11$

which simplifies to

$16 + 2 b = 11$ $16+2b=11$

$\Rightarrow 2 b = - 5$ $=>2b=-5$

$\Rightarrow b = - \frac{5}{2}$ $=>color(blue)(b=-5/2)$

Now, we've solved for both of our variables.

Hope this helps!