Question 1095782
THE FIFTH-GRADER SOLUTION:
With all the answers right, a student would get
{{{2*60=120}}} marks.
Each wrong answer makes a student lose {{{2+1=3}}} marks,
2 because he did not choose a right answer,
and 1 more because he chose a wrong answer.
A student answered every question and lost {{{120-90=30}}} of the possible marks
must have {{{"30 ÷ 3"=10}}} wrong answers.
He answered the other {{{60-10=50}}} questions correctly.
 
THE ALGEBRA CLASS SOLUTION:
{{{r}}}= number of right answers
{{{w}}}= number of wrong answersin
As all questions were answered,
{{{r+w=60}}} .
The marks for tthe test are
{{{2r-w=90}}}
 
Then, you have to solve the system of linear equations {{{system(r+w=60,2r-w=90)}}} .
There are many ways to do that.
 
COMBINATIONS:
You could do
{{{2r-w=90}}}
{{{"+"}}}
{{{" "}}}{{{r+w=60}}}
{{{"---------"}}}
{{{3r}}} {{{"="}}}{{{150}}} ---> {{{"r = 150 ÷ 3"}}} --->{{{r=50}}}
Then you could just substitute that value for {{{r}}} in {{{r+w=60}}}
to get {{{50+w=60}}} ---> {{{w=60-50}}} ---> {{{w=10}}} .
Otherwise, you could keep doing combinations of equations,
and multiply
{{{2r-w=90}}} times (-1) to get {{{-2r+w=-90}}} ,
{{{r+e=60}}} times 2 to get {{{2r+2w=1200}}} ,
and add them up to get
{{{-2r+w=-90}}}
{{{"+"}}}
{{{2r+2w=120}}}
{{{"___________"}}}
{{{" "}}} {{{" "}}} {{{" "}}}{{{3w=30}}} ---> {{{"w = 30 ÷ 3"}}} ---> {{{w=10}}}
Teachers call that adding up of equations a "method" to solve systems of linear equations by making making combinations of the equations, often calling it the "combinations method." Your teacher may use another name.
Why it works and is reasonable:
If {{{r+w}}} and {{{60}}} are the same number,
and adding the same number to both sides of {{{2r-w=90}}} we get an equivalent equation,
we can add {{{r+w}}} to the left side and {{{60}}} to the right side,
to get the "combination" equation
{{{3r=90+60}}} .
We could also multiply both sides of {{{r+w=60}}} times (-2) and add them to {{{2r-w=90}}}
 
SUBSTITUTION:
What teachers call "substitution method"
requires solving one of the equations for one of the variables,
and then substituting the expression found into the other equation.
You choose which equation to solve and which variable to solve for.
There are 4 possible choices.
One of them is shown below.
 
{{{2r-w=90}}} {{{2r=90+w}}} {{{2r-90=w}}}
 
{{{system(r+w=60,w=2r-90)}}} {{{r+2r-90=60}}} {{{3r-90=60}}}{{{3r=60+90}}}{{{3r=150}}}{{{r=150/3}}}{{{r=50}}}