Solve each system using substitution. 3x+2y=23 1/2x-4=y

2 answers:

5 0

$\left\{\begin{array}{ccc}3x+2y=23&(1)\\y=\frac{1}{2}x-4&(2)\end{array}\right\\\\subtitute\ (2)\ to\ (1)\\\\3x+2(\frac{1}{2}x-4)=23\\3x+x-8=23\\4x-8=23\ \ \ \ |add\ 8\ to\ both\ sides\\4x=31\ \ \ \ \ |divide\ both\ sides\ by\ 4\\\boxed{x=7.75}\\\\subtitute\ the\ value\ of\ "x"\ to\ (2)\\\\y=\frac{1}{2}(7.75)-4\\y=3.875-4\\\boxed{y=-0.125}\\\\Answer:\boxed{ \left\[\begin{array}{ccc}x=7.75\ (7\frac{3}{4})\\\\y=-0.125\ (-\frac{1}{8})\end{array}\right}$

Harman [31]4 years ago

5 0

1: 3x+2y=23
2: 1/2x-4=y
We have that y=1/2x-4 in 2
So lets substitu y in 1
3x+2(1/2x-4)=23
3x+x-8=23
4x=23+8
4x=31
X = 7,75
Now lets replace x in 2:
1/2.7,75-4=y
3,875-4 + y
Y= - 0,125
So (x;y)=(7,75;-0.125)

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Rudiy27

Answer:

Adding the 2 equations.

Step-by-step explanation:

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solmaris [256]

Answer: It's a tie between f(x) and h(x). Both have the same max of y = 3

The highest point shown on the graph of f(x) is at (x,y) = (pi,3). The y value here is y = 3.

For h(x), the max occurs when cosine is at its largest: when cos(x) = 1.
So,
h(x) = 2*cos(x)+1
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showing that h(x) maxes out at y = 3 as well

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