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Zorluk: ZorLinear Equations in One Variable

In the equation below, kk is a constant.

13(2kx−9)−56(x+4)=112\frac{1}{3}(2kx - 9) - \frac{5}{6}(x + 4) = \frac{11}{2}

If the equation has no solution, what is the value of kk?

Cevap: 1.25

Cevap

1.25 (or 5/4)
The correct answer is 1.25 (or 5/4). A linear equation in one variable of the form Ax+B=CAx + B = C has no solution if the variable terms on both sides of the equation are equal (meaning A=0A = 0) and the constant terms are unequal (B≠CB \neq C). Expanding the left side of the given equation yields 23kx−3−56x−103=112\frac{2}{3}kx - 3 - \frac{5}{6}x - \frac{10}{3} = \frac{11}{2}. Combining the constant terms gives (23k−56)x−193=112\left(\frac{2}{3}k - \frac{5}{6}\right)x - \frac{19}{3} = \frac{11}{2}. Setting the coefficient of xx to 00 yields 23k−56=0\frac{2}{3}k - \frac{5}{6} = 0. Solving for kk gives k=56×32=54k = \frac{5}{6} \times \frac{3}{2} = \frac{5}{4}, which is equivalent to 1.25. Since the remaining constant terms are unequal (−193≠112-\frac{19}{3} \neq \frac{11}{2}), the equation has no solution when k=1.25k = 1.25.

Adım Adım Çözüm

1
Expand the expression on the left side of the equation
23kx−3−56x−103=112\frac{2}{3}kx - 3 - \frac{5}{6}x - \frac{10}{3} = \frac{11}{2}
Apply the distributive property to remove the parentheses.
2
Group the xx terms and combine the constants on the left side
(23k−56)x−193=112\left(\frac{2}{3}k - \frac{5}{6}\right)x - \frac{19}{3} = \frac{11}{2}
Simplify the equation by combining like terms: −3−103=−93−103=−193-3 - \frac{10}{3} = -\frac{9}{3} - \frac{10}{3} = -\frac{19}{3}.
3
Set the coefficient of the xx term equal to 00
23k−56=0\frac{2}{3}k - \frac{5}{6} = 0
For a linear equation to have no solution, the variable terms on both sides of the equation must cancel out (meaning the coefficient of the variable must be 00), while the remaining constant terms must not be equal (−193≠112-\frac{19}{3} \neq \frac{11}{2}).
4
Solve the resulting equation for kk
k=1.25k = 1.25
Add 56\frac{5}{6} to both sides to get 23k=56\frac{2}{3}k = \frac{5}{6}, then multiply by the reciprocal of 23\frac{2}{3}, which gives k=56×32=1512=54=1.25k = \frac{5}{6} \times \frac{3}{2} = \frac{15}{12} = \frac{5}{4} = 1.25.

Anahtar Kavram

Conditions for a linear equation in one variable to have no solution
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Konular

Linear Equations in One Variable

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