Question.
Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9, 3x + 2y = 4
Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9, 3x + 2y = 4
Solution:
By substitution method,
8x + 5y = 9 ...(i)
3x + 2y = 4 ...(ii)
From (ii), we get
$x=\frac{4-2 y}{3}$
Substituting x from (iii) in (i), we get
$8\left(\frac{4-2 y}{3}\right)+5 y=9$
= 32 – 16y + 15y = 27
= 5 = y
Substituting y = 5 in (ii) we get
3x + 2(v) = 4
= 3x = – 6
= x = – 2
Hence, x = – 2, y = 5
By cross multiplication method
8x + 5y = 9 ...(i)
3x + 2y = 4 ...(ii)
$\frac{x}{5 \times(-4)-2(-9)}=\frac{y}{3 \times(-9)-8(-4)}=\frac{1}{8 \times 2-3 \times 5}$
$\Rightarrow x=\frac{-\mathbf{2 0}+\mathbf{1 8}}{\mathbf{1}}=-2$
$\Rightarrow y=\frac{-\mathbf{2 7}+\mathbf{3 2}}{\mathbf{1}}=5$
Hence, x = –2, y = 5
By substitution method,
8x + 5y = 9 ...(i)
3x + 2y = 4 ...(ii)
From (ii), we get
$x=\frac{4-2 y}{3}$
Substituting x from (iii) in (i), we get
$8\left(\frac{4-2 y}{3}\right)+5 y=9$
= 32 – 16y + 15y = 27
= 5 = y
Substituting y = 5 in (ii) we get
3x + 2(v) = 4
= 3x = – 6
= x = – 2
Hence, x = – 2, y = 5
By cross multiplication method
8x + 5y = 9 ...(i)
3x + 2y = 4 ...(ii)
$\frac{x}{5 \times(-4)-2(-9)}=\frac{y}{3 \times(-9)-8(-4)}=\frac{1}{8 \times 2-3 \times 5}$
$\Rightarrow x=\frac{-\mathbf{2 0}+\mathbf{1 8}}{\mathbf{1}}=-2$
$\Rightarrow y=\frac{-\mathbf{2 7}+\mathbf{3 2}}{\mathbf{1}}=5$
Hence, x = –2, y = 5
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