I have a question about solving a linear equation by substitution.
"solve each system by substitution"
4x + 5y = 7
9y = 31 + 2x
When neither unknown has a coefficient of 1, -1 or 0 in either equation, the
system is best done by elimination, not substitution. But since you were told
to do it by substitution, here goes.
If there were any fractions you would clear of them. There are no fractions,
so:
Pick the equation and letter whose coefficient is smallest in absolute value.
Solve that equation for that letter:
Of 4x, 5y, 9y and 2x, 2x has the smallest coefficient in absolute value, so
choose the second equation and solve it for x
9y = 31 + 2x
9y - 31 = 2x
x = (9y - 31)/2
Substitute (9y - 3)/2 in the equation that has not yet been used:
4x + 5y = 7
4[(9y - 31)/2] + 5y = 7
Cancel the 2 into the 4:
2
4[(9y - 31)/2] + 5y = 7
1
2(9y - 31) + 5y = 7
18y - 62 + 5y = 7
23y - 62 = 7
23y = 69
y = 3
Now substitute 3 for y in
x = (9y - 31)/2
x = (9·3 - 31)/2
x = (27 - 31)/2
x = -4/2
x = -2
Edwin
AnlytcPhil@aol.com
