Answer:
3x=-y+12
2x+3y=-6
collecting the like terms we have
3x+y=12
2x+3y=-6
to solve this simultaneous equation, we use substitution method
we make y the subject in equation one:
y=12-3x
then we put y=12-3x into 2x-3y=-6 in equation two
2x+3(12-3x)=-6
2x+36-9x=-6
-7x+36=-6
-7x=-6-36
-7x=-42
x=-42÷-7
x=6
then we put x=6 in y=12-3x:
y=12-3x
y=12-3(6)
y=12-18
y=-6
Answer:
Step-by-step explanation:
(8x²-18x+10)/(x²+5)(x-3)
express the expression as a partial fraction:
(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +bx+c/x²+5
both denominator are equal , so require only work with the nominator
(8x²-18x+10)=(x²+5)A+(x-3)(bx+c)
8x²-18x+10= x²A+5A+bx²+cx-3bx-3c
combine like terms:
x²(A+b)+x(-3b+c)+5A-3c
(8x²-18x+10)
looking at the equation
A+b=8
-3b+c=-18
5A-3c=10
solve for A,b and c (system of equation)
A=2 , B=6, and C=0
substitute in the value of A, b and c
(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +(bx+c)/x²+5
(8x²-18x+10)/[(x^2+5)(x-3)] = 2/x-3 + (6x+0)/(x²+5)
(8x²-18x+10)/[(x^2+5)(x-3)] =
<h2>2/(x-3)+6x/x²+5</h2>
(4x+2)/[(x²+4)(x-2)]
(4x+2)/[(x²+4)(x-2)]= A/(x-2) + bx+c/(x²-2)
(4x+2)=a(x²-2)+(bx+c)(x-2)
follow the same step in the previous answer:
the answer is :
<h2>(4x+2)/[(x²+4)(x-2)]= 5/4/(x-2) + (3/2 -5x/4)/(x²+4)</h2>
Need more information or a graphic.
Pi and 4, because the square root of ten is 3.1622...
1) 3+2+15=20
2) 3/4= 27/36 1/9=4/36
3) 27/36+4/36=31/36
4) 26/27=936/972 31/36= 637/972
5) 936/972 + 637/972= 1573/972= 1 601/972
