2, 4, 6, 8, 10, 12
difference is 2
223
-119
104
In stead of doing 3-9 and 2-1 do 23-19 then 2-1
Answer:y= -4
Step-by-step explanation:
1 Solve for xx in 7x-4y=-127x−4y=−12.
x=\frac{4(y-3)}{7}
x=
7
4(y−3)
2 Substitute x=\frac{4(y-3)}{7}x=
7
4(y−3)
into 9x-4y=-209x−4y=−20.
\frac{36(y-3)}{7}-4y=-20
7
36(y−3)
−4y=−20
3 Solve for yy in \frac{36(y-3)}{7}-4y=-20
7
36(y−3)
−4y=−20.
y=-4
y=−4
4 Substitute y=-4y=−4 into x=\frac{4(y-3)}{7}x=
7
4(y−3)
.
x=-4
x=−4
5 Therefore,
\begin{aligned}&x=-4\\&y=-4\end{aligned}
x=−4
y=−4
The median is 79. Hope this helps!
Suppose we wish to determine whether or not two given polynomials with complex coefficients have a common root. Given two first-degree polynomials a0 + a1x and b0 + b1x, we seek a single value of x such that
Solving each of these equations for x we get x = -a0/a1 and x = -b0/b1 respectively, so in order for both equations to be satisfied simultaneously we must have a0/a1 = b0/b1, which can also be written as a0b1 - a1b0 = 0. Formally we can regard this system as two linear equations in the two quantities x0 and x1, and write them in matrix form as
Hence a non-trivial solution requires the vanishing of the determinant of the coefficient matrix, which again gives a0b1 - a1b0 = 0.
Now consider two polynomials of degree 2. In this case we seek a single value of x such that
Hope this helped, Hope I did not make it to complated
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