How do you determine whether a linear system has one solution, many solutions, or no solution when given 8x =+ y = -16 and -3x + y = - 5?

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Answer 1 · 2018-04-10T20:42:43

There is one solution. $x=-1$ and $y=-8$

I'm assuming the two equations are

Equation I: $8x+y=-16$ , and

Equation II: $-3x+y=-5$ .

I will subtract Equation II from Equation I.

$8x+y-(-3x)-y=-16-(-5)$

Which simplifies to

$11x=-11$ .

Divide both sides by 11.

$x=-1$

Substitute $x=-1$ into Equation II.

$-3(-1)+y=-5$ .

Simplify.

$3+y=-5$

Subtract three from both sides of this equation.

$y=-8$

There is one solution. $x=-1$ and $y=-8$