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Natali5045456 [20]
3 years ago
10

How do I solve this equation?

4" id="TexFormula1" title="x + 8y = 20 \\ 2x + 4y = 4" alt="x + 8y = 20 \\ 2x + 4y = 4" align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
bixtya [17]3 years ago
7 0
X = -8y + 20
2(-8y + 20) +4y = 4
-16y + 40 + 4y = 4
-12y +40=4
Subtract 40 from each side
-12y= -36
Divide both by -12
y = 3
Plug Y=3 in to find x
X + 8(3) = 20
X + 24 = 20
Subtract 24
x = -4
ArbitrLikvidat [17]3 years ago
6 0

Answer:

Subtract x x from both sides of the equation. 8y=20−x 8 y = 20 - x. Divide each term by 8 8 and simplify.

Solve for x 2x-4y=4. 2x−4y=4 2 x - 4 y = 4. Add 4y 4 y to both sides of the equation. 2x=4+4y 2 x = 4 + 4 y. Divide each term by 2 2 and simplify.

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(3^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=27\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=-x^2+11x+26

or if you mean (3x^3+5x^2+3x-7)+(8x-6x^2+6):

(3x^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=3x^3\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=3x^3-x^2+11x-1

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(8x-4x^2+3)-(x^3+7x^2+3x-8)=\\\\=\underline{8x}\ \underline{\underline{-\ 4x^2}}+3-x^3\ \underline{\underline{-\,7x^2}}\ \underline{-\,3x}+8=\\\\=-x^3-11x^2+5x+10

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