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Zorluk: ZorLinear Equations in Two Variables

In the xyxy-plane, the graph of the linear equation ax+by=cax + by = c, where aa, bb, and cc are constants, has a slope of 23\frac{2}{3} and passes through the point (6,5)(6, 5). If a+b=5a + b = 5, what is the value of cc?

Cevap: 15

Cevap

15
Rearranging the linear equation ax+by=cax + by = c into slope-intercept form y=abx+cby = -\frac{a}{b}x + \frac{c}{b} shows the slope is ab-\frac{a}{b}. Setting ab=23-\frac{a}{b} = \frac{2}{3} gives 2b=3a2b = -3a, or b=1.5ab = -1.5a. Substituting this into a+b=5a + b = 5 yields a1.5a=5a - 1.5a = 5, which simplifies to 0.5a=5-0.5a = 5, so a=10a = -10. This means b=1.5(10)=15b = -1.5(-10) = 15. Substituting the values of a=10a = -10 and b=15b = 15 along with the point (6,5)(6, 5) into the equation ax+by=cax + by = c yields (10)(6)+(15)(5)=15(-10)(6) + (15)(5) = 15, so the value of cc is 1515.

Adım Adım Çözüm

1
Rewrite the standard form equation ax+by=cax + by = c in slope-intercept form.
y=abx+cby = -\frac{a}{b}x + \frac{c}{b}
This allows the identification of the slope of the line in terms of the coefficients aa and bb.
2
Set the slope expression equal to the given slope of 23\frac{2}{3} and solve for bb in terms of aa.
b=1.5ab = -1.5a
The slope of the line is ab-\frac{a}{b}, so ab=23    2b=3a    b=1.5a-\frac{a}{b} = \frac{2}{3} \implies 2b = -3a \implies b = -1.5a.
3
Substitute b=1.5ab = -1.5a into the given equation a+b=5a + b = 5 and solve for aa, then find bb.
a=10a = -10 and b=15b = 15
Substituting gives a1.5a=5    0.5a=5    a=10a - 1.5a = 5 \implies -0.5a = 5 \implies a = -10. Substituting a=10a = -10 back into the relationship gives b=1.5(10)=15b = -1.5(-10) = 15.
4
Substitute a=10a = -10, b=15b = 15, and the point (6,5)(6, 5) into the equation ax+by=cax + by = c and solve for cc.
c=15c = 15
(10)(6)+(15)(5)=c    60+75=15    c=15(-10)(6) + (15)(5) = c \implies -60 + 75 = 15 \implies c = 15.

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Linear Equations in Two Variables
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