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Lostsunrise [7]
3 years ago
15

4x+y=8 and x+3y=8 graphed

Mathematics
1 answer:
IRISSAK [1]3 years ago
3 0
standard \ linear \ equation\ :\\\\y=ax+b\\\\ \begin{cases} 4x+y=8\ \ |\ subtract\ 4x\ to\ both\ sides\\ x+3y=8 \ \ |\ subtract\ x\ to\ both\ sides \end{cases}\\\\\begin{cases} y=-4x+8 \\ 3y=-x+8\ \ | \ divide \ each \ term \ by \ 3 \end{cases}

\begin{cases} y=-4x+8 \\ y=-\frac{1}{3}x+\frac{8}{3} \end{cases}\\\\y=-4x+8\\ To \ find \ the \ x-axis \ intersection \ point, \\set \ y \ equal \ to \ zero \ and \ solve \ for \ x : \\ \\y=0 \ \to 0=-4x+8\\\\4x=8 \ \ | \ divide \ both \ sides\ by\ 4 \\\\x=2\\\\ point : \ \ (2,0)
 
To \ find \ the \ y-axis \ intersection \ point, \\set \ x \ equal \ to \ zero \ and \ solve \ for \ y : \\ \\x=0 \ \to y=-4 \cdot 0+8\\ y=8 \\ point: \ \ (0,8)


y=-\frac{1}{3}x+\frac{8}{3}\\\\ \ the \ x-axis \ intersection \ point \\ \\y=0 \ \to 0=-\frac{1}{3}x+\frac{8}{3}\\ \frac{1}{3}x=\frac{8}{3} \ \ | \ multiply\ both\ sides\ by\ 3 \\\\x=8 \\\\point: \ \ (8,0)

the \ y-axis \ intersection \ point \\ \\x=0 \ \to y=-\frac{1}{3} \cdot 0+\frac{8}{3} \\ y=\frac{8}{3} \\ point : \ \ (0,\frac{8}{3})


Answer :\\\\ \begin{cases} y=-4x+8 \ \ | \ multiply \ each \ term \ by \ (-3) \\ 3y=-x+8 \end{cases}\\\begin{cases} -3y=12x-24 \\ 3y=-x+8 \end{cases}\\+-------\\0=11x-16\\11x=16\ \ | \ divide \ both \ sides\ by\ 11\\x=\frac{16}{11}

y=-4 \cdot \frac{16}{11}+8 \\ y=- \frac{ 64}{11}+\frac{88}{11}\\y= \frac{24}{11}\\\\\begin{cases} x=\frac{16}{11} \\ y=\frac{24}{11} \end{cases}


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8z=4(2z+1) show step by step
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8z=4(2z+1)
First you would distribute the constant into the numbers in the parentheses. so
8z=8z+4
then you would combine like terms which in this case would result in a zero.
8z-8z=0 So the z would be zero or no solution.
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Find the coordinates of point z such that the ratio of fz to zd is 5;3
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The point such that the coordinate is 5;3 is (14, 0)

<h3>Midpoint of coordinates using ratio</h3>

The formula for finding the midpoint of a line in the ratio m:n is expressed as:

M(x, y) = {(mx₁+nx₂)/2, (my₁+ny₂)/2,}

Given the coordinate of G and D on the line as G(5, 0) and D(1,0)

Since there is no y-axis, hence;

x = 5(5)+1(3)/2

x = 25+3/2

x = 28/2

x =14

Hence the point such that the ratio is 5;3 is (14, 0)

Learn more on midpoint of a line here: brainly.com/question/5566419

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