Hi Andrea;
1. y=x+5
y=4x1
y=y
Let's substitute...
x+5=4x1
Let's isolate x...
Let's subtract 5 from both sides...
5+x+5=4x15
x=4x6
Let's subtract 4x from both sides...
4x+x=4x64x
3x=6
Let's divide both sides by 3...
(3x)/3=6/3
x=2
Let's plug the value of x into the first equation to establish the value of y...
y=x+5
y=2+5
y=7
Let's plug the value of x into the second equation to establish the value of y and verify the results of the first equation...
y=4x1
y=4(2)1
y=81
y=7
Results match!
2. y=x/2
y = x4
y=y
Let's substitute...
x/2=x4
Let's eliminate the denominator, also known as the divisor, of 2 by dividing both sides by 2...
2(x/2)=2(x4)
x=2x8
Let's subtract 2x from both sides as we proceed to isolate x...
2x+x=2x82x
x=8
Let's multiply both sides by 1...
(1)(x)=8(1)
x=8
Let's plug this value of x into the first equation and solve for y...
y=x/2
y=8/2
y=4
Let's plug this value of x into the second equation and solve for y to verify the results of the first...
y=x4
y=84=4
The results match!
3. x+y=5
2x3y=5
Let's isolate y using either equation. The first is the easiest...
x+y=5
y=5x
2x3y=5
Let's substitute y for (5x)...
(2x)[3(5x)]=5
2x15+3x=5
Let's combine like terms...
5x15=5
Let's add 15 to both sides as we proceed to isolate x...
15+5x15=5+15
5x=20
Let's divide both sides by 5...
(5x)/5=20/5
x=4
Let's plug this value of x into the first equation to solve for y...
x+y=5
4+y=5
y=1
Let's plug this value of x into the second equation to solve for y and verify the results of the first equation...
2x3y=5
2(4)3y=5
83y=5
3y=3
y=1
The results match.
10/30/2013

Vivian L.
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