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Zorluk: OrtaSystems of Linear Equations

Consider the system of linear equations below:

3x+4y=8ax+8y=15\begin{aligned} 3x + 4y &= 8 \\ ax + 8y &= 15 \end{aligned}

If the system of equations has no solution, and aa is a constant, what is the value of aa?

Cevap: 6

Cevap

The correct answer is 6.
For a system of linear equations to have no solution, the lines representing the equations must be parallel, which requires them to have the same slope but different y-intercepts. Writing 3x+4y=83x + 4y = 8 in slope-intercept form gives y=34x+2y = -\frac{3}{4}x + 2, so its slope is 34-\frac{3}{4}. Writing ax+8y=15ax + 8y = 15 in slope-intercept form gives y=a8x+158y = -\frac{a}{8}x + \frac{15}{8}, so its slope is a8-\frac{a}{8}. Equating the slopes gives 34=a8-\frac{3}{4} = -\frac{a}{8}. Solving for aa gives a=6a = 6. Since the y-intercepts (22 and 158\frac{15}{8}) are different, the lines are parallel and have no intersection points.

Adım Adım Çözüm

1
Convert the first equation 3x+4y=83x + 4y = 8 to slope-intercept form.
y=34x+2y = -\frac{3}{4}x + 2
To find the slope of the first line, which is 34-\frac{3}{4}.
2
Convert the second equation ax+8y=15ax + 8y = 15 to slope-intercept form.
y=a8x+158y = -\frac{a}{8}x + \frac{15}{8}
To find the slope of the second line in terms of aa, which is a8-\frac{a}{8}.
3
Equate the slopes of the two lines.
34=a8-\frac{3}{4} = -\frac{a}{8}
Parallel lines have the same slope, and a system with parallel lines has no solution.
4
Solve the equation for aa.
a=6a = 6
Isolating the variable aa yields the value of the constant.

Anahtar Kavram

A system of linear equations has no solution if the lines represented by the equations are parallel, meaning they have the same slope but different y-intercepts.
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