PLEASE HELP I FORGOT HOW TO DO THIS

Mathematics

1 answer:

kozerog [31]3 years ago

7 0

If I’m not mistaken your answer would be 240

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X-y=4 and 1/2x+1/2y=18

cluponka [151]

value of x = 20 and value of y=16

Solution set=(20,16)

Step-by-step explanation:

We need to solve system of equations:

$x-y=4\\\frac{1}{2}x+ \frac{1}{2}y=18$

Let:

$x-y=4\,\,\,eq(1)\\\frac{1}{2}x+ \frac{1}{2}y=18\,\,\,eq(2)$

Simplifying equation 2:

$\frac{1}{2}x+ \frac{1}{2}y=18$

Taking LCM on left side:

$\frac{x+y}{2}=18$

$x+y=18*2$

$x+y=36\,\,\,eq(3)$

Adding eq(1) and eq(3)

$x-y=4\\x+y=36\\-----\\2x=40\\x=40/2\\x=20$

Putting value of x in eq(1) and finding y:

$x-y=4\\20-y=4\\-y=4-20\\-y=-16\\y=16$

So, value of x = 20 and value of y=16

Solution set=(20,16)

Keywords: System of equations

Learn more about system of equations at:

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8 0

4 years ago

Find the fifth term of (2x + y)^10

mamaluj [8]

$T_r={n \choose r-1}a^{n-(r-1)}b^{r-1}\\T_5={10 \choose 5-1}(2x)^{10-(5-1)}y^{5-1}\\T_5={10 \choose 4}(2x)^6y^{4}\\T_5=\frac{10!}{4!6!}64x^6y^{4}\\T_5=\frac{7\cdot8\cdot9\cdot10}{2\cdot3\cdot4}64x^6y^{4}\\T_5=7\cdot3\cdot10\cdot64x^6y^{4}\\T_5=13440x^6y^4$