<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B50x-2y%3D230%7D%20%5Catop%20%7B50x-3y%3D170%7D%20%5Cright." id="TexFo

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3 years ago

rmula1" title="\left \{ {{50x-2y=230} \atop {50x-3y=170} \right." alt="\left \{ {{50x-2y=230} \atop {50x-3y=170} \right." align="absmiddle" class="latex-formula">

combination method

Mathematics

1 answer:

grin007 [14]3 years ago

5 0

Answer:

x = 7, y = 60

Step-by-step explanation:

Given the 2 equations

50x - 2y = 230 → (1)

50x - 3y = 170 → (2)

Subtract (2) from (1) term by term on both sides

y = 60

Substitute y = 60 into either of the 2 equations and evaluate for x

Substituting into (1)

50x - 2(60) = 230

50x - 120 = 230 ( add 120 to both sides )

50x = 350 ( divide both sides by 50 )

x = 7

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Approximate the integral integral integral integral f(x, y) dA by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2),

amm1812

Answer:

Step-by-step explanation:

Approximate the integral $\int\int\limits_R {f(x,y)} \, dA$ by dividing the region $R$ with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.

Find the sum $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i$

Since all are equal squares, so $\delta A_i=1$ for every $i$

$\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=f(x_1,y_1)\delta A_1+f(x_2,y_2)\delta A_2+f(x_3,y_3)\delta A_3+f(x_4,y_4)\delta A_4+f(x_5,y_5)\delta A_5+f(x_6,y_6)\delta A_6+f(x_7,y_7)\delta A_7+f(x_8,y_8)\delta A_8\\\\=f(0.5,0.5)(1)+f(1.5,0.5)(1)+f(2.5,0.5)(1)+f(3.5,0.5)(1)+f(0.5,1.5)(1)+f(1.5,1.5)(1)+f(2.5,1.5)(1)+f(3.5,1.5)(1)\\\\=0.5+0.5+1.5+0.5+2.5+0.5+3.5+0.5+0.5+1.5+1.5+1.5+2.5+1.5+3.5+1.5\\\\=24$

Thus, $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=24$

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