The answer would be B. You would need to keep the first fraction, change to multiplication, and flip the second fraction and multiply across.
If you're asking for the model for the second problem (I didn't see any unsolved), here it is!
An equvilent equation
remember you can do anything to an equation as long asyou do it to both sides
assuming yo have
x+y=1 and
x-3y=9
mulitply both by 2
2x+2y=2
2x-6y=18
those are equvilent
ok, solve initial
x+y=1
x-3y=9
multiply first equation by -1 and add to 2nd equation
-x-y=-1
<u>x-3y=9 +</u>
0x-4y=8
-4y=8
divide both sides by -4
y=-2
sub back
x+y=1
x-2=1
add 2
x=3
x=3
y=-2
(3,-2)
if we test it in other one
2x+2y=2
2(3)+2(-2)=2
6-4=2
2=2
yep
2x-6y=18
2(3)-6(-2)=18
6+12=18
18=18
yep
solution is (3,-2)
Answer:
Sheila have 6 bills
Step-by-step explanation:
Suppose Sheila has x '20' dollar bills, so the total cost of her bill = $20x
Suppose Tamsin has y '10' dollar bills, so the total cost of her bill = $10y
<h3>Equation 1</h3>
<em>Sheila has a total of $50 more than Tamsin</em>
<h2><em>10y + 50 = 20x</em></h2><h2 /><h3>
Equation 2</h3>
<em>Tamsin has 1 more bill than Sheila.</em>
<h2>y = x + 1 </h2><h2 />
<em>Now solve them by putting value of x in equation </em>
10( x + 1 ) + 50 = 20x
10x + 10 + 50 = 20x
60 = 20x - 10x
10x = 60
x = 60/10
x = 6
and
y = 7
<em> </em>
