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Solve each system by substitution. If a system has no solution or infinitely many solutions, so state.$$\left\{\begin{array}{l}{a-3 b=-1} \\{-b=-2 a-2}\end{array}\right.$$
$b=0$$a=-1 $
Algebra
Chapter 4
Systems of Linear Equations and Inequalities
Section 2
Solving Systems of Equations by Substitution
Equations and Inequalities
Systems of Equations and Inequalities
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have a minus three b is minus one and minus B his minus two. A minus four minus two. So let's let's change the second equation here by dividing everything by negative one. So we will get B is to a plus two across this one out. Going to use be here to plug into the first equation. So a minus three into to a plus two is equal to negative one a minus six a minus six is minus one. So we're gonna have minus five. A minus six is negative one. Now let's add six on both sides minus five. A is five divide both sides by negative five will get a is negative one. Now, since we know that B is to a plus two. B equals two A plus two that's using this equation and plug in a here. So be is to into negative one plus two. So B is minus two plus two. So be is zero
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