Question

Difficulty: HardLinear 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?

Answer: 15

Answer

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 a−1.5a=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.

Step-by-Step Solution

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 a−1.5a=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.

Key Concept

Linear Equations in Two Variables
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