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Zorluk: ZorSolving Linear Equations

For what value of aa does the equation 34(x−2)−13(2x+a)=112x−5\frac{3}{4}(x - 2) - \frac{1}{3}(2x + a) = \frac{1}{12}x - 5 have infinitely many solutions for xx?

Cevap: 10.5

Cevap

10.5
Expanding the left side of the equation yields 34x−32−23x−a3\frac{3}{4}x - \frac{3}{2} - \frac{2}{3}x - \frac{a}{3}. Combining the coefficients of xx gives (34−23)x=112x(\frac{3}{4} - \frac{2}{3})x = \frac{1}{12}x. The equation becomes 112x−(32+a3)=112x−5\frac{1}{12}x - (\frac{3}{2} + \frac{a}{3}) = \frac{1}{12}x - 5. For a linear equation to have infinitely many solutions, the variable coefficients must be equal and the constant terms must also be equal. Therefore, we equate the constants: −32−a3=−5-\frac{3}{2} - \frac{a}{3} = -5. Multiplying all terms by −6-6 to clear the denominators yields 9+2a=309 + 2a = 30, which simplifies to 2a=212a = 21, or a=10.5a = 10.5.

Adım Adım Çözüm

1
Expand and simplify the left side of the equation by distributing the fraction coefficients.
112x−32−a3\frac{1}{12}x - \frac{3}{2} - \frac{a}{3}
Distributing 34\frac{3}{4} and −13-\frac{1}{3} across their respective parentheses and combining the xx terms allows us to compare the coefficients on both sides.
2
Equate the constant terms from both sides of the equation.
−32−a3=−5-\frac{3}{2} - \frac{a}{3} = -5
A linear equation of the form Ax+B=Cx+DAx + B = Cx + D has infinitely many solutions if and only if A=CA = C and B=DB = D. Since both AA and CC are 112\frac{1}{12}, we set the constant terms equal.
3
Isolate the variable aa and solve.
a=10.5a = 10.5
Adding 32\frac{3}{2} to both sides gives −a3=−3.5-\frac{a}{3} = -3.5. Multiplying both sides by −3-3 yields the final value.

Anahtar Kavram

Solving linear equations with infinitely many solutions by equating coefficients and constant terms.
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Solving Linear Equations

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