Question
Mathematics
$\begin{array}{l}5a-6b-3c=4\\ 4a-2b-4c=-6\\ -2a+6b-3c=-22\end{array}${:[5a-6b-3c=4],[4a-2b-4c=-6],[-2a+6b-3c=-22]:}
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To solve this system of equations, we can use the method of elimination or substitution. Here, I will show you the elimination method.
First, let’s eliminate the variable $a$a. We can do this by multiplying the second equation by $5$5 and the third equation by $-1$-1, and then adding them to the first equation. This will eliminate $a$a:
$5\left(4a-2b-4c\right)+\left(-1\right)\left(-2a+6b-3c\right)=5\left(-6\right)+\left(-1\right)\left(-22\right)$5(4a-2b-4c)+(-1)(-2a+6b-3c)=5(-6)+(-1)(-22)
Simplifying, we get:
$20a-10b-20c+2a-6b+3c=-30+22$20 a-10 b-20 c+2a-6b+3c=-30+22
Combining like terms, we have:
$22a-16b-17c=-8$22 a-16 b-17 c=-8
Next, let’s eliminate the variable $b$b. We can do this by multiplying the first equation by $2$2 and the third equation by $3$3, then adding them to the second equation:
$2\left(5a-6b-3c\right)+3\left(-2a+6b-3c\right)=2\left(4\right)+3\left(-2\right)$2(5a-6b-3c)+3(-2a+6b-3c)=2(4)+3(-2)
Simplifying, we get:
$10a-12b-6c-6a+18b-9c=8-6$10 a-12 b-6c-6a+18 b-9c=8-6
Combining like terms, we have:
$4a+6b-15c=2$4a+6b-15 c=2
Now we have two equations:
$22a-16b-17c=-8$22 a-16 b-17 c=-8
$4a+6b-15c=2$4a+6b-15 c=2
To eliminate the variable $b$b, let’s multiply the first equation by $3$3 and the second equation by $8$8, then subtract the equations:
$3\left(22a-16b-17c\right)-8\left(4a+6b-15c\right)=-8-16$3(22 a-16 b-17 c)-8(4a+6b-15 c)=-8-16
Simplifying, we get:
$66a-48b-51c-32a-48b+120c=-24-16$66 a-48 b-51 c-32 a-48 b+120 c=-24-16
Combining like terms, we have:
$34a-96b+69c=-40$34 a-96 b+69 c=-40
Now we have two equations:
$34a-96b+69c=-40$34 a-96 b+69 c=-40
$4a+6b-15c=2$4a+6b-15 c=2
To eliminate the variable $a$a, let’s multiply the second equation by $17$17 and the first equation by $4$4, then subtract the equations:
$17\left(4a+6b-15c\right)-4\left(34a-96b+69c\right)=2-\left(-40\right)$17(4a+6b-15 c)-4(34 a-96 b+69 c)=2-(-40)
Simplifying, we get:
$68a+102b-255c-136a+384b-276c=42+80$68 a+102 b-255 c-136 a+384 b-276 c=42+80
Combining like terms, we have:
$-68a+486b-531c=122$-68 a+486 b-531 c=122
Now we have two equations:
$34a-96b+69c=-40$34 a-96 b+69 c=-40
$-68a+486b-531c=122$-68 a+486 b-531 c=122
We can solve for $a$a in the first equation:
$a=\frac{-96b+69c-40}{34}$a=(-96 b+69 c-40)/(34)
Substituting this into the second equation, we get:
$-68\left(\frac{-96b+69c-40}{34}\right)+486b-531c=122$-68((-96 b+69 c-40)/(34))+486 b-531 c=122
Simplifying this expression, we get:
$96b-69c-80+486b-531c=122$96 b-69 c-80+486 b-531 c=122
Combining like terms, we have:
$582b-600c=202$582 b-600 c=202
To solve for $b$b, we can divide this equation by $582$582:
$b=\frac{600c+202}{582}$b=(600 c+202)/(582)
Finally, we can substitute the values of $a$a and $b$b back into any of the original equations to solve for $c$c. Let’s use the first equation:
$5a-6b-3c=4$5a-6b-3c=4
After simplifying and solving for $c$c, we find:
$c=\frac{415}{582}$c=(415)/(582)
Substituting this value back into the equation for $b$b, we get:
$b=\frac{203}{291}$b=(203)/(291)
And substituting the values of $b$b and $c$c back into the equation for $a$a, we get:
$a=\frac{-115}{291}$a=(-115)/(291)
Therefore, the solution to the system of equations is:
$a=\frac{-115}{291}$a=(-115)/(291), $b=\frac{203}{291}$b=(203)/(291), $c=\frac{415}{582}$c=(415)/(582)
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