Video: Computing Numerical Expressions Using Factorisation

If 3𝑝 + 2π‘ž = 4 and 6π‘Ž βˆ’ 5𝑏 = 10, what is the value of 20(π‘ž + 𝑏) + 6(5𝑝 βˆ’ 4π‘Ž)?

03:42

Video Transcript

If three 𝑝 plus two π‘ž equals four and six π‘Ž minus five 𝑏 equals 10, what is the value of 20 times π‘ž plus 𝑏 plus six times five 𝑝 minus four π‘Ž?

Let’s first write down what we know. We know that three 𝑝 plus two π‘ž equals four and that six π‘Ž minus five 𝑏 equals 10. What we want to do now is compare the grouping of variables for our expression in the two equations. In our first equation, 𝑝 and π‘ž are grouped together. In our second equation, π‘Ž and 𝑏 are grouped together. But when we look at 𝑝 and π‘ž in the expression we’re trying to solve, they are not grouped together, nor our π‘Ž and 𝑏. And that means we want to take this expression and try to regroup it so that 𝑝 and π‘ž can be grouped together and π‘Ž and 𝑏 can be grouped together. In order to do that, we’ll need to distribute this multiplication by 20 and the multiplication by six across the parentheses.

20 times π‘ž is 20π‘ž. 20 times 𝑏 is 20𝑏. Six times five 𝑝 is 30𝑝. Six times negative four π‘Ž is negative 24π‘Ž. Because at this point, we’re only dealing with addition and subtraction, we can regroup. So we’ll gather the 𝑝 and π‘ž together: 30𝑝 plus 20π‘ž. Again, in this step, the only thing we’re doing is rearranging the equation. We’re not changing its value. Now, we’ll group the π‘Ž and 𝑏: negative 24π‘Ž plus 20𝑏. At this point, we want to try to rewrite 30𝑝 plus 20π‘ž to be in the format some constant 𝑐 times three 𝑝 plus two π‘ž. If we take out a factor of 10, we haven’t changed the value because 10 times three 𝑝 equals 30𝑝 and 10 times two π‘ž equals 20π‘ž. And so, for now, we’ll just leave that part of the expression as 10 times three 𝑝 plus two π‘ž.

We want to do the same thing with our final two terms. Is there some constant 𝑐 we can remove so that the expression is 𝑐 times six π‘Ž minus five 𝑏? We need the coefficient with π‘Ž to become six instead of negative 24. Negative four times six π‘Ž equals negative 24π‘Ž. And if we take out a factor of negative four, we need to divide 20 by negative four to find the coefficient four 𝑏, which will be negative five. Negative four times negative five 𝑏 equals positive 20𝑏.

After all of this rearranging, we can do some substitution. We know that three 𝑝 plus two π‘ž equals four. And so, in place of three 𝑝 plus two π‘ž, we substitute four. We know that six π‘Ž minus five 𝑏 equals 10. And so in place of six π‘Ž minus five 𝑏, we substitute 10 and bring down the multiplied by negative four. We now have something that says 10 times four minus four times 10. And so, we can say that 40 minus 40 equals zero.

The value of 20 times π‘ž plus 𝑏 plus six times five 𝑝 minus four π‘Ž is zero.

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