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Zorluk: ZorLinear Equations in Two Variables

The table below shows several values of xx and their corresponding values of yy for a linear relationship, where pp and kk are constants.

xxyy
pp44
p+2p + 21010
2p+32p + 3kk

If k=25k = 25, what is the value of pp?

  1. A
    0.5
  2. 4Cevap
  3. C
    6
  4. D
    60

Cevap

4
The correct answer is 4. A linear relationship has a constant rate of change (slope). Calculating the slope mm from the first two points (p,4)(p, 4) and (p+2,10)(p + 2, 10) gives m=104(p+2)p=3m = \frac{10 - 4}{(p + 2) - p} = 3. The equation of the line is y4=3(xp)y - 4 = 3(x - p), which simplifies to y=3x3p+4y = 3x - 3p + 4. Substituting the third point (2p+3,25)(2p + 3, 25) into this equation gives 25=3(2p+3)3p+425 = 3(2p + 3) - 3p + 4, which simplifies to 25=3p+1325 = 3p + 13. Solving for pp yields p=4p = 4.

Adım Adım Çözüm

1
Determine the slope of the linear relationship using the first two points from the table, (p,4)(p, 4) and (p+2,10)(p + 2, 10).
The slope is m=3m = 3.
A linear relationship has a constant rate of change (slope), which is the change in yy divided by the change in xx: m=104(p+2)p=62=3m = \frac{10 - 4}{(p + 2) - p} = \frac{6}{2} = 3.
2
Write the equation of the line using the point-slope form with the point (p,4)(p, 4) and slope m=3m = 3.
y=3x3p+4y = 3x - 3p + 4
Expressing the linear relationship as an equation allows us to find the relationship between the parameters pp and kk: y4=3(xp)y=3x3p+4y - 4 = 3(x - p) \Rightarrow y = 3x - 3p + 4.
3
Substitute the third point (2p+3,k)(2p + 3, k) into the linear equation and simplify the expression to solve for kk in terms of pp.
k=3p+13k = 3p + 13
Since the point (2p+3,k)(2p + 3, k) lies on the line, its coordinates must satisfy the line's equation: k=3(2p+3)3p+4k=6p+93p+4k=3p+13k = 3(2p + 3) - 3p + 4 \Rightarrow k = 6p + 9 - 3p + 4 \Rightarrow k = 3p + 13.
4
Substitute the given value k=25k = 25 into the equation k=3p+13k = 3p + 13 and solve for pp.
p=4p = 4
This isolates the variable pp to find its numerical value under the specified conditions: 25=3p+1312=3pp=425 = 3p + 13 \Rightarrow 12 = 3p \Rightarrow p = 4.

Anahtar Kavram

Finding the equation of a line from a table of values and using substitution to solve for unknown parameters in a linear relationship.

Alternatif Yöntem

An alternative method is to use the property that the slope between any two points on a line is constant. We can equate the slope between (p,4)(p, 4) and (p+2,10)(p + 2, 10) to the slope between (p,4)(p, 4) and (2p+3,25)(2p + 3, 25). This gives the equation 104(p+2)p=254(2p+3)p\frac{10 - 4}{(p + 2) - p} = \frac{25 - 4}{(2p + 3) - p}, which simplifies to 3=21p+33 = \frac{21}{p + 3}. Solving this equation yields 3(p+3)=213(p + 3) = 21, so p+3=7p + 3 = 7, which gives p=4p = 4.
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