<span>Multiply one of the equations so that both equations share a common complementary coefficient.
In order to solve using the elimination method, you need to have a matching coefficient that will cancel out a variable when you add the equations together. For the 2 equations given, you have a huge number of choices. I'll just mention a few of them.
You can multiply the 1st equation by -2/5 to allow cancelling the a term.
You can multiply the 1st equation by 5/3 to allow cancelling the b term.
You can multiply the 2nd equation by -2.5 to allow cancelling the a term.
You can multiply the 2nd equation by 3/5 to allow cancelling the b term.
You can even multiply both equations.
For instance, multiply the 1st equation by 5 and the second by 3. And in fact, let's do that.
5a + 3b = –9
2a – 5b = –16
5*(5a + 3b = -9) = 25a + 15b = -45
3*(2a - 5b = -16) = 6a - 15b = -48
Then add the equations
25a + 15b = -45
6a - 15b = -48
=
31a = -93
a = -3
And then plug in the discovered value of a into one of the original equations and solve for b.</span>
Answer:
The equation has infinitely many solutions for any value of P and Q such that P=Q.
Step-by-step explanation:
Px - 37 = Qx - 37
Px - Qx - 37 = -37
x (P-Q) = 0
==> P-Q=0 ==> P=Q
Answer:
The location of the point is between Quadrant II and Quadrant III
Step-by-step explanation:
we know that
The abscissa refers to the x-axis and ordinate refers to the y-axis
so
in this problem we have
the coordinates of the point are ![(-x,0)](https://tex.z-dn.net/?f=%28-x%2C0%29)
see the attached figure to better understand the problem
The location of the point is between Quadrant II and Quadrant III
Answer:
![(n+2)^2+6](https://tex.z-dn.net/?f=%28n%2B2%29%5E2%2B6)
Step-by-step explanation:
Pattern 1 consists of
small squares (here n = 1 and
).
Pattern 2 consists of
small squares (here n = 2 and
).
Pattern 3 consists of
small squares (here n = 3 and
).
Thus, pattern n consists of
small squares.
Answer:
f(-2)=-2
Step-by-step explanation:
we know that
x -----> is the independent variable or input value
f(x) ---> is the dependent variable or output value
f(-2) is the value of the function f(x) when the value of x is equal to -2
so
For x=-2
Looking at the graph
f(-2)=-2
For an input value of -2 the output value is -2