 hexacordoK

2021-02-08

Find values of a and b such that the system of linear equations has (a) no solution, (b) exactly one solution, and (c) infinitely many solutions.
x + 2y = 3
ax + by = −9 irwchh

Given,
The system of linear equations, x + 2y = 3, ax + by = −9.
Part a)
To find values of a,b such that given system has no solution:
Linear equations with no solution are inconsistent equation or the graph of such equations do not intersect that is these lines are parallel.
Consider the first equation,
$x+2y=3$
$⇒2y=-x+3$
$⇒y=-\frac{x}{2}+3$
Above equation has slope $-\frac{1}{2}$.
Consider another equation,
$ax+by=-9$
$⇒by+=ax=9$
$⇒y=\frac{ax}{b}-\frac{9}{b}$
Slope of above equation is $-\frac{a}{b}$.
As given system of equations have no solution the given equations are parallel lines and they have same slopes.
That is,
$-\frac{a}{b}=-\frac{1}{2}$
$⇒a=1,b=2$
Thus, for a = 1 and b = 2 , the given system has no solution.
Part b)
Given system of equations has exactly one solution.
Linear equations with exactly one solution are lines which intersect at exactly one point.
Also, intersecting lines have negative reciprocal slopes.
From part a), first equation have slope $-\frac{1}{2}$ and its negative reciprocal is 2.
Also, slope of second equation is $-\frac{a}{b}$ .
Therefore,
$-\frac{a}{b}=\frac{2}{1}$
$⇒a=-2,b=1$
Algebra homework question answer, step 3, image 1
Thus, for a = -2 and b = 1 or a = 2 and b = -1 the given system has exactly one solution.
.
Part c)
Given system of equation has infinitely many solutions,
Linear equations with infinitely many solutions are lines that intersect at all points.
They are multiples of original equations.
To solve for a and b, multiple of the first equation's constant with the second equation's constant , Just multiply it to the coefficients of the first equations variable.
For multiple,
$-\frac{9}{3}=-3$
Then, multiply first equation by -3,
$-3\left(x+2y=3\right)$
$⇒-3x-6y=-9$
$⇒a=-3,b=-6$
Thus, for a = -3 and b = -6 the system of linear equations have infinitely many solutions.

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