Question #bab68

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Answer 1 · 2018-01-18T05:41:13

$26$ math problems and $16$ science problems.

Let :

So

$x+y=42$
$x=y+10$

Then substitute $y+10$ in for $x$ to solve for $y$ :

$y+10+y=42$

$2y+10=42$

$2y=32$

$y=16$

Plug $y$ into either equation to solve for $x$ :

$x=16+10$

$x=26$

Answer 2 · 2018-01-18T06:04:04

There are $26$ math problems and $16$ science problems.

Let $x$ equal the number of science problems, and $x+10$ equal the number of math problems. Set their sum equal to $42$ . Solve for $x$ .

$x+x+10=42$

Simplify.

$2x+10=42$

Subtract $10$ from both sides.

$2x+10-10=42-10$

Simplify.

$2x=32$

Divide both sides by $2$ .

$x=32/2$

Simplify.

$x=16$

Number of science questions: $x=16$

Number of math questions: $x+10=16+10=26$