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sleet_krkn [62]
3 years ago
14

Apply Crammer's Rule to find the solution to the following quations .2x + 3y = 1, 3x + y = 5​

Mathematics
1 answer:
Bond [772]3 years ago
4 0

Answer:

The solution to the equation system given is:

  • <u>x = 2</u>
  • <u>y = -1</u>

Step-by-step explanation:

First, we must know the equations given:

  1. 2x + 3y = 1
  2. 3x + y = 5​

Following Crammer's Rule, we have the matrix form:

\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] =\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]

Now we solve using the determinants:

x=\frac{\left[\begin{array}{ccc}1&3\\5&1\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(1*1)-(5*3)}{(2*1)-(3*3)} = \frac{1-15}{2-9} =\frac{-14}{-7} = 2

y=\frac{\left[\begin{array}{ccc}2&1\\3&5\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(2*5)-(3*1)}{(2*1)-(3*3)}=\frac{10-3}{2-9} =\frac{7}{-7}=-1

Now, we can find the answer which is x= 2 and y= -1, we can replace these values in the equation to confirm the results are right, with the first equation:

  • 2x + 3y = 1
  • 2(2) + 3(-1)= 1
  • 4 - 3 = 1
  • 1 = 1

And, with the second equation:

  • 3x + y = 5​
  • 3(2) + (-1) = 5
  • 6 - 1 = 5
  • 5 = 5

 

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2 years ago
How many real number roots does the equation have? Use the discriminant. 3x^2-6x+3=0
tresset_1 [31]
We know that
in a quadratic equation of the form
ax²+bx+c=0
the discriminant is
D=b²-4ac
if D> 0 <span>then there are two real number solutions
if D=0 </span><span>then there's only one real number solution.
if D< 0 </span><span>then there are two complex number solutions

in this problem
we have
</span>3x²<span>-6x+3=0
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so 
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therefore
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4 0
3 years ago
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