Examkin

Soru

Zorluk: Çok zorLinear Equations in One Variable
In the equation below, aa and bb are positive constants.
23(32x−a)−34(8−bx)=52(x−3)−12\frac{2}{3} \left( \frac{3}{2}x - a \right) - \frac{3}{4} \left( 8 - bx \right) = \frac{5}{2}(x - 3) - \frac{1}{2}
If the equation has infinitely many solutions, what is the value of aba^b?

Cevap: 9

Cevap

The correct answer is 9.
The correct answer is 9. Expanding the left side of the equation yields (1+34b)x−(6+23a)\left(1 + \frac{3}{4}b\right)x - \left(6 + \frac{2}{3}a\right), and simplifying the right side yields 52x−8\frac{5}{2}x - 8. For the linear equation to have infinitely many solutions, the coefficient of xx on the left, 1+34b1 + \frac{3}{4}b, must equal the coefficient of xx on the right, 52\frac{5}{2}, which gives b=2b = 2. Similarly, the constant term on the left, −(6+23a)-\left(6 + \frac{2}{3}a\right), must equal the constant term on the right, −8-8, which simplifies to 6+23a=86 + \frac{2}{3}a = 8 and gives a=3a = 3. Evaluating aba^b with these values yields 32=93^2 = 9.

Adım Adım Çözüm

1
Expand both sides of the equation to collect like terms.
(1+34b)x−(6+23a)=52x−8\left( 1 + \frac{3}{4}b \right)x - \left( 6 + \frac{2}{3}a \right) = \frac{5}{2}x - 8
Expanding allows us to compare the coefficient of xx and the constant term on each side of the equation.
2
Equate the coefficients of xx on both sides of the equation.
1+34b=52  ⟹  b=21 + \frac{3}{4}b = \frac{5}{2} \implies b = 2
For a linear equation to have infinitely many solutions, the coefficient of xx must be identical on both sides.
3
Equate the constant terms on both sides of the equation.
−(6+23a)=−8  ⟹  a=3-\left( 6 + \frac{2}{3}a \right) = -8 \implies a = 3
For a linear equation to have infinitely many solutions, the constant terms must also be identical on both sides.
4
Calculate the value of aba^b using the solved values of aa and bb.
32=93^2 = 9
The question asks for the value of the expression aba^b where a=3a = 3 and b=2b = 2.

Anahtar Kavram

A linear equation in one variable 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.

Alternatif Yöntem

Since the equation must hold for all values of xx if it has infinitely many solutions, you can substitute convenient values for xx to solve for aa and bb directly. Substituting x=0x = 0 simplifies the equation to −23a−6=−8-\frac{2}{3}a - 6 = -8, which quickly yields a=3a = 3. Then, substituting x=2x = 2 and a=3a = 3 simplifies the equation to −34(8−2b)=−3-\frac{3}{4}(8 - 2b) = -3, which yields b=2b = 2. Calculating aba^b gives 32=93^2 = 9.
Tahmini Süre:3m 0s
Bu soruyu puanla

Konular

Linear Equations in One Variable

Bu konudan daha fazla soru